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Nicole throws a ball straight up. Chad watches the ball from a window \(5.0 \mathrm{m}\) above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of \(10 \mathrm{m} / \mathrm{s}\) as it passes him on the way back down. How fast did Nicole throw the ball?

Short Answer

Expert verified
Nicole threw the ball at approximately 0.078 m/s at Chad's level.

Step by step solution

01

Apply the formula to find the time taken for the ball to fall from the peak to Chad's level

We use the following formula describing the relationship between displacement \(\Delta y\), initial velocity \(V_i\), acceleration \(a\) and time \(t\): \(\Delta y = V_i t + 0.5 a t^2\). Since the ball is at its peak height when it starts to fall down, implying its initial velocity is zero at that point, the formula becomes: \(\Delta y = 0.5 a t^2\). Substituting the values \(\Delta y = 5.0 m\) and \(a = -9.8 m/s^2\), the equation can be rearranged to solve for \(t\): \(t = \sqrt{(-2*\Delta y / a)}\)
02

Calculate time taken to fall

Substituting the given values in our rearranged equation gives: \(t = \sqrt{(-2*-5.0 / 9.8)} = 1.01 s\). This is the time taken for the ball to fall from its highest point to Chad's height.
03

Use the time to find the initial velocity

We can use this time \(t\) to find the initial speed of the ball as it passes Chad's level on the way upwards, since as previously mentioned, the time taken to reach maximum height will be equivalent to the time taken to fall from maximum height to Chad's level. We use the equation representing the relationship between final velocity \(V_f\), initial velocity \(V_i\), acceleration \(a\) and time \(t\): \(V_f = V_i + a*t\). Since the final velocity when it reaches Chad is -10 m/s (it is negative because it is moving downwards now), and using the acceleration due to gravity of -9.8 m/s^2, we can substitute into the rearranged form of the equation \(V_i = V_f - a * t\)
04

Calculate initial velocity

Substituting the given values into the equation gives: \(V_i = -10 - (-9.8) * 1.01 = 0.078 m/s\). Hence, the speed at which Nicole threw the ball when it was at Chad's level was approximately 0.078 m/s upwards.

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Key Concepts

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

Kinematics in Projectile Motion
Kinematics forms the basis for understanding how objects move under various forces. It's particularly important in projectile motion, where we study the path taken by objects thrown or propelled into the air. In this scenario, we focus on several key variables:
  • Displacement (\( \Delta y \))
  • Velocity
  • Acceleration
  • Time
Each variable helps us describe the motion of an object, breaking down complex movements into simpler parts.
Using kinematics, we can predict and calculate the position of an object at a given time, the speed at different intervals, and understand how acceleration affects the object's path.
In projectile motion like Nicole's ball throw, gravity acts as constant acceleration downward at 9.8 \( m/s^2 \). This force changes the velocity of the ball throughout its path.
Calculating Initial Velocity
To find the initial velocity, we apply equations of motion that establish a relationship between the variables mentioned earlier. The equation used in this case is:\[ V_f = V_i + a \cdot t \]where:
  • \( V_f \) is the final velocity
  • \( V_i \) is the initial velocity
  • \( a \) is the acceleration (gravity in this case)
  • \( t \) is the time
For Nicole's ball, we want to calculate how fast she threw the ball upwards initially. As the ball comes back to Chad's level, \( V_f \) is known (10 \( m/s \) downward). By rearranging the formula to solve for \( V_i \), we substitute the collected values to determine how quickly the ball was moving as it left her hand. This calculation considers the effect of gravity during the ascent and descent of the projectile.
Understanding Free Fall
Free fall describes any motion of a body where gravity is the only force acting upon it, giving it an acceleration of 9.8 \( m/s^2 \) downwards. In the problem, after the initial throw, the ball is in a state of free fall, especially as it falls back towards the ground. The ball initially moves upwards until the force of gravity slows it, stops it momentarily at the peak, and then accelerates it back downwards. Here are some important points about free fall:
  • All objects in free fall near earth surface accelerate downwards at the same rate, regardless of mass.
  • The initial upward motion is slowed down by gravity until the velocity becomes zero at the peak.
  • The subsequent downward motion increases in speed as gravity continues to accelerate it.
Observing how free fall operates in this type of motion helps in understanding energy conservation, as kinetic energy transforms to potential energy and back.

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Most popular questions from this chapter

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within \(10 \mathrm{cm}\) of the end of the ramp without going off. You give the puck a push, releasing it with a speed of \(5.0 \mathrm{m} / \mathrm{s}\) when it is \(8.5 \mathrm{m}\) from the end of the ramp. The puck's speed after traveling \(3.0 \mathrm{m}\) is \(4.0 \mathrm{m} / \mathrm{s}\). Are you a winner?

A car starts at the origin and moves with velocity \(\vec{v}=\) \((10 \mathrm{m} / \mathrm{s},\) northeast). How far from the origin will the car be after traveling for 45 s?

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?

A particle's velocity is described by the function \(v_{x}=k t^{2} \mathrm{m} / \mathrm{s}\) where \(k\) is a constant and \(t\) is in \(s .\) The particle's position at \(t_{0}=0 \mathrm{s}\) is \(x_{0}=-9.0 \mathrm{m} .\) At \(t_{1}=3.0 \mathrm{s},\) the particle is at \(x_{1}=9.0 \mathrm{m} .\) Determine the value of the constant \(k .\) Be sure to include the proper units.

The position of a particle is given by the function \(x=\) \(\left(2 t^{3}-9 t^{2}+12\right) \mathrm{m},\) where \(t\) is in \(\mathrm{s}\) a. At what time or times is \(v_{x}=0 \mathrm{m} / \mathrm{s} ?\) b. What are the particle's position and its acceleration at this \(\operatorname{time}(s) ?\)

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