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Chapter 2: Problem 12

From the top of a tower of height \(50 \mathrm{~m}\), a ball is thrown vertically upwards with a certain velocity. It hits the ground \(10 \mathrm{~s}\) after it is thrown up. How much time does it take to cover a distance \(A B\) where \(A\) and \(B\) are two points \(20 \mathrm{~m}\) and \(40 \mathrm{~m}\) below the edge of the tower? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(2.0 \mathrm{~s}\) (b) \(1.0 \mathrm{~s}\) (c) \(0.5 \mathrm{~s}\) (d) \(0.4 \mathrm{~s}\)

Short Answer

Expert verified
Option (d) 0.4 s.

Step by step solution

01

Identify Initial Conditions

The initial height of the ball is at the top of the tower, which is \(50\, \text{m}\) above the ground. Let's denote the initial velocity of the ball as \(u\). The ball's motion is influenced by gravity \(g = 10 \, \text{ms}^{-2}\). It takes \(10 \,\text{s}\) to hit the ground after being thrown upwards.
02

Determine Initial Velocity

Using the equation for the motion under gravity:\[s = ut + \frac{1}{2}gt^2\]where \(s\) is the displacement from the top to the ground, which is \(-50\, \text{m}\), since it's falling downwards eventually. We substitute \(t = 10 \, \text{s}\), \(g = -10 \, \text{ms}^{-2}\):\[-50 = 10u + \frac{1}{2}(-10)(10^2)\]\[-50 = 10u - 500\]\[450 = 10u\]\[u = 45\, \text{ms}^{-1}\].Now we know the initial velocity is \(45 \, \text{ms}^{-1}\).
03

Calculate Time to Reach Point A and B

Now, let's find the times at which the ball is at points \(A\) and \(B\). We again use the same motion equation:\[s = ut + \frac{1}{2}gt^2\]For point \(A\) where \(s = -20\, \text{m}\):\[-20 = 45t - 5t^2\]Rearranging, we get:\[5t^2 - 45t - 20 = 0\] This can be solved using the quadratic formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 5\), \(b = -45\), \(c = -20\): \[t = \frac{45 \pm \sqrt{(-45)^2 - 4 \cdot 5 \cdot (-20)}}{2 \cdot 5}\]\[t = \frac{45 \pm \sqrt{2025 + 400}}{10}\]\[t = \frac{45 \pm \sqrt{2425}}{10}\]\[t = \frac{45 \pm 49.25}{10}\]\[t_A = \frac{45 + 49.25}{10} \approx 9.425 \text{s}\] (Ignoring second root for now)\[t'_A = \frac{45 - 49.25}{10} \approx -0.425 \text{s}\] (Neglect as a valid time to reach as it's negative).
04

Repeat for Point B

For point \(B\) where \(s = -40\, \text{m}\):\[-40 = 45t - 5t^2\]Rearranging gives:\[5t^2 - 45t - 40 = 0\]Using the quadratic formula again, where \(a = 5\), \(b = -45\), \(c = -40\):\[t = \frac{45 \pm \sqrt{(-45)^2 - 4 \cdot 5 \cdot (-40)}}{10}\]\[t = \frac{45 \pm \sqrt{2025 + 800}}{10}\]\[t = \frac{45 \pm \sqrt{2825}}{10}\]\[t_B = \frac{45 + 53.16}{10} \approx 9.816 \text{s}\] (Choose this root as time is positive)\[t'_B = \frac{45 - 53.16}{10} \approx -0.816 \text{s}\] (Again negative, discard it as valid time).
05

Calculate Time Interval Between A and B

The time interval it takes to travel between points \(A\) and \(B\) is:\[\Delta t = t_B - t_A\]\[\Delta t = 9.816\, \text{s} - 9.425\, \text{s}\]\[\Delta t \approx 0.4\, \text{s}\].
06

Determine Correct Option

The calculated time taken \(\Delta t\) matches with option (d). Therefore, the correct answer is (d) 0.4 s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Motion
Vertical motion describes the movement of an object under the influence of gravity alone. It mainly involves two key phases:
  • Upward Motion: When the object is thrown upwards, it moves against gravity. This phase is characterized by a gradual decrease in velocity until it reaches its peak, where the velocity becomes zero.
  • Downward Motion: After reaching the peak, gravity causes the object to accelerate downwards until it hits the ground. This phase involves increasing velocity.
In the context of the problem, the ball is thrown vertically from the top of a 50 m tower. The upward motion will have an initial velocity which we calculated as 45 m/s, and the gravitational pull of 10 m/s虏 will slow it down, eventually reversing its direction. Hence, understanding vertical motion helps grasp how the object travels and how long it takes to hit particular points as it moves up and down the path.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation of the form:\[ ax^2 + bx + c = 0 \]
  • In our exercise, we used the quadratic equation to find the times when the ball reaches certain points below the release height.
  • By substituting the known values of displacement, initial velocity, and acceleration due to gravity into the equation, we get \[ 5t^2 - 45t - C = 0 \]where \(C\) is the corrected variable for different points along the trajectory (20 m or 40 m below the tower).
Solving the quadratic equation can be done using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The roots obtained represent the possible times at which the projectile reaches a particular height. By calculating the roots, we determine the time intervals necessary for the projectile to travel between given points.
Displacement Calculation
Displacement calculation involves determining the change in position of an object, factoring in direction and magnitude.
  • In our case, given two points (20 m and 40 m below the tower's edge), we calculate how long it takes the ball to move between these points as it travels in vertical motion.
  • The formula typically used is \[ s = ut + \frac{1}{2}gt^2 \]Here \(s\) is the displacement, \(u\) the initial velocity, \(t\) the time, and \(g\) the acceleration due to gravity.
  • The displacement calculations help ascertain how long it takes for the projectile to pass a certain vertical distance after being subjected to the initial upward velocity and gravity.
By accurately calculating displacement, students can determine key factors influencing motion such as speed, time, and the effect of gravity on vertical motion.

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Most popular questions from this chapter

A stone thrown vertically upwards attains a maximum height of \(45 \mathrm{~m} .\) In what time the velocity of stone become equal to one-half the velocity of throw? (Given \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(2 \mathrm{~s}\) (b) \(1.5 \mathrm{~s}\) (c) \(1 \mathrm{~s}\) (d) \(0.5 \mathrm{~s}\)

At a metro station, a girl walks up a stationary escalator in time \(t_{1}\). If she remains stationary on the escalator, then the escalator take her up in time \(t_{2}\). The time taken by her to walk up on the moving escalator will be [NCERT Exemplar] (a) \(\left(t_{1}+t_{2}\right) / 2\) (b) \(t_{1} t_{2} /\left(t_{2}-t_{1}\right)\) (c) \(t_{1} t_{2} /\left(t_{2}+t_{1}\right)\) (d) \(t_{1}-t_{2}\)

The acceleration of a particle is increasing linearly with time \(t\) as \(b t\). The particle starts from the origin with an initial velocity \(v_{0}\). The distance travelled by the particle in time \(t\), is (a) \(v_{0} t+\frac{1}{3} b t^{2}\) (b) \(v_{0} t+\frac{1}{6} b t^{3}\) (c) \(v_{0} f+\frac{1}{3} b t^{3}\) (d) \(v_{0} t+\frac{1}{2} b t^{2}\)

On a two lane road, car \(A\) is travelling with a speed of \(36 \mathrm{~km} / \mathrm{h}\). Two cars \(B\) and \(C\) approach car \(A\) in opposite directions with a speed of \(54 \mathrm{~km} / \mathrm{h}\) each. At a certain instant, when the distance \(A B\) is equal to \(A C\), both being \(1 \mathrm{~km}, B\) decides to overtake \(A\) before \(C\) does. In this case, the acceleration of car \(B\) is required to avoid an accident (a) \(1 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(0.1 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(1.9 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(0.2 \mathrm{~m} / \mathrm{s}^{2}\)

An express train is moving with a velocity \(v_{1}\) its driver finds another train is moving on the same track in the same direction with velocity \(v_{2}\). To avoid collision driver applies a retardation \(a\) on the train. The minimum time of avoiding collision will be (a) \(t=\frac{v_{1}-v_{2}}{a}\) (b) \(t=\frac{v_{1}^{2}-v_{2}^{2}}{2}\) (c) None (d) Both (a) and (b)
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