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Chapter 2: Problem 30

A ball is thrown vertically upwards from the ground. It crosses a point at the height of \(25 \mathrm{~m}\) twice at an interval of \(4 \mathrm{~s}\). The ball was thrown with the velocity of \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(20 \mathrm{~m} / \mathrm{s}\) (B) \(25 \mathrm{~m} / \mathrm{s}\) (C) \(30 \mathrm{~m} / \mathrm{s}\) (D) \(35 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The initial velocity of the ball thrown vertically upwards is approximately \(22.5 \mathrm{~m/s}\), which is closest to option (B) \(25 \mathrm{~m/s}\).

Step by step solution

01

List the known information

The following information is known: 1. The height of the ball above the ground: \(h = 25\) m 2. The time interval to cross the height twice: \(\Delta t = 4\) s 3. The acceleration due to gravity: \(g = 10\) m/s虏, acting downwards
02

Understand that the ball reaches the given height twice

The ball reaches the 25 m mark twice: once when it is moving upwards, and once again when it is moving downwards. Therefore, the 4 s time interval account for both the ascent and descent time. As the motion is symmetrical, we can divide the time interval by 2 to find the time to reach 25 m while going up: \[t_{up} = \frac{\Delta t}{2} = \frac{4}{2} = 2 \mathrm{s}\]
03

Use the kinematic equation for height in terms of time

We will use the following kinematic equation for the height (h) reached by an object under constant acceleration (g): \[h = v_{0}t - \frac{1}{2}gt^2\] Here, \(v_{0}\) is the initial velocity we're trying to find. We know the height (h), time (t), and acceleration (g). We can plug in these values into the equation and solve for the initial velocity.
04

Substitute values and solve for the initial velocity

Substitute the known values (h = 25 m, t = 2 s, g = 10 m/s虏) into the kinematic equation: \[25 = v_{0}(2) - \frac{1}{2}(10)(2)^2\] Now, simplify the equation to find the initial velocity, \(v_{0}\): \[25 = 2v_{0} - 20\] \[v_{0} = \frac{25 + 20}{2} = \frac{45}{2} = 22.5 \mathrm{~m/s}\]
05

Choose the answer closest to the calculated initial velocity

The calculated initial velocity is 22.5 m/s. The given options are: (A) 20 m/s (B) 25 m/s (C) 30 m/s (D) 35 m/s The option closest to our calculated value is (B) \(25 \mathrm{~m/s}\). Therefore, the answer is (B).

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Key Concepts

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

Projectile Motion
Projectile motion refers to the motion of an object that is thrown or projected into the air, subject to only the force of gravity. In the case of our exercise, the ball follows a trajectory or path that is parabolic due to this gravitational force.

Here are the key aspects:
  • Trajectory: The ball moves upward, reaches a peak height, and then descends back to the ground.
  • Dimensional Motion: While projectiles can move in two dimensions, this problem simplifies it to vertical (one-dimensional) motion.
  • Symmetry: The projectile's path is symmetrical in time. The time taken to reach a certain height is the same on the way up as it is on the way down.

In our situation, after being thrown upwards, the ball reaches the 25 m mark on its way up, and again on its way down. The problem is to determine the initial velocity required for this symmetrical motion.
Initial Velocity Calculation
Initial velocity is vital in determining how far and how high a projectile will travel. It is the speed at which the ball is launched upwards. To find it, we use the kinematic equation:

\[ h = v_{0}t - \frac{1}{2}gt^2 \] Where:
  • \( h \) is height.
  • \( v_{0} \) is the initial velocity we want to find.
  • \( t \) is the time taken to reach the height.
  • \( g \) is the acceleration due to gravity.
By plugging in known values (\( h = 25 \) m, \( t = 2 \) s, \( g = 10 \) m/s虏), we solved for \( v_{0} \) and found it to be 22.5 m/s. This calculation helps us determine the plausible answer from multiple-choice options given in the exercise.
Acceleration Due to Gravity
Acceleration due to gravity is a constant force that acts on all objects near the Earth's surface. In physics, it is denoted as \( g \) and typically considered to be \( 9.8 \) m/s虏 or approximated to \( 10 \) m/s虏 for easier calculations in exercises like this one.

Key elements about gravity:
  • Influence: It pulls objects towards the Earth's center, causing the deceleration of a projectile going upwards and acceleration as it comes down.
  • Uniform Value: The consistent value of \( g = 10 \) m/s虏 was used here, simplifying our calculations.
  • Role in Motion: It dictates the projectile鈥檚 rise and fall, affecting both time and distance traveled.
In this exercise, the gravity's uniform acceleration helps predict the motion of the ball and enables us to calculate how initial velocity is counteracted by the gravitational pull during its ascent.

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Most popular questions from this chapter

In a harbour, wind is blowing at the speed of \(72 \mathrm{~km} / \mathrm{h}\) and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of \(51 \mathrm{~km} / \mathrm{h}\) to the north, what is the direction of the flag on the mast of the boat? (A) \(\tan ^{-1} \frac{51}{72 \sqrt{2}-51}\) (B) \(\tan ^{-1} \frac{72 \sqrt{2}-51}{51}\) (C) \(\tan ^{-1} 1\) (D) None

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A particle moving along a straight line with uniform acceleration displaces by \(13 \mathrm{~m}\) and \(7 \mathrm{~m}\) in second and fifth second of its motion respectively. Column-I (A) Distance (in \(\mathrm{m}\) ) travelled in first \(9 \mathrm{~s}\) (B) Magnitude of acceleration (in \(\mathrm{ms}^{-2}\) ) of particle (C) Displacement (in \(\mathrm{m}\) ) in first \(7 \mathrm{~s}\) (D) Time (in s) when displacement is zero Column-II (1) 63 (2) 2 (3) 65 (4) 16

The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is (A) \(\theta\) (B) \(\tan ^{-1}(\theta / t)\) (C) \(\tan ^{-1}\left(\frac{v \cos \theta}{v \sin \theta-g t}\right)\) (D) \(\tan ^{-1}\left(\frac{v \sin \theta-g t}{v \cos \theta}\right)\)

A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes \(3 \mathrm{~s}\) to pass the passenger, and the last boggy takes \(2 \mathrm{~s}\) to pass him. If passenger is late by \(t\) second from the departure of the train, then the find the value of \(2 \mathrm{t}\). (Assume that the train accelerates at constant rate and all the boggies are of equal length.)
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