Questions: A ball is launched upward from the ground with initial velocity of 14 m / s and reaches height h above the ground before falling back down. Take the upward direction to be positive. Refer to the figure. Neglect air resistance. - Part (a) What is the acceleration, in meters per second squared, of the ball when it is in the air? a=-9.800 m / s^2 - Part (b) What is the velocity of the ball, in meters per second, when it reaches the top, or highest point, of its trajectory? Yp=0.000 m / s - Part (c) Enter an expression for the height of the ball as a function of time in terms of the initial velocity y1 and the acceleration a and the elapsed time r. h(t)=v1 t+1 / 2 a t^2 - Part (d) What is the maximum height the ball reaches in meters? hmix=1000 m Part (e) Enter an expression for the elapsed time it takes for the ball to travel from the ground to a given height in terms of the initial velocity vi, the velocity the ball has at that height m, and the acceleration a. Δt=

A ball is launched upward from the ground with initial velocity of 14 m / s and reaches height h above the ground before falling back down. Take the upward direction to be positive. Refer to the figure. Neglect air resistance. - Part (a) What is the acceleration, in meters per second squared, of the ball when it is in the air? a=-9.800 m / s^2 - Part (b) What is the velocity of the ball, in meters per second, when it reaches the top, or highest point, of its trajectory? Yp=0.000 m / s - Part (c) Enter an expression for the height of the ball as a function of time in terms of the initial velocity y1 and the acceleration a and the elapsed time r. h(t)=v1 t+1 / 2 a t^2 - Part (d) What is the maximum height the ball reaches in meters? hmix=1000 m Part (e) Enter an expression for the elapsed time it takes for the ball to travel from the ground to a given height in terms of the initial velocity vi, the velocity the ball has at that height m, and the acceleration a. Δt=
Transcript text: A ball is launched upward from the ground with initial velocity of $14 \mathrm{~m} / \mathrm{s}$ and reaches height $h$ above the ground before falling back down. Thke the upward direction to be positive. Refer to the figure. Neglect air resistance. ? - Part (a) $V$ What is the acceleration, in meters per second squared, of the ball when it is in the air? \[ \begin{array}{l} a=-9.800 \mathrm{~m} / \mathrm{s}^{2} \end{array} \] - Part (b) $V$ What is the velocity of the ball, in meters per second, when it reaches the top, or highest point, of its trajectory? \[ Y_{\mathrm{p}}=0.000 \mathrm{~m} / \mathrm{s} \] - Part (c) $V$ Enter an expression for the height of the ball as a function of time in terms of the initial velocity $y_{1}$ and the acceleration $a$ and the elapsed time $r$. \[ h(t)=v_{1} t+1 / 2 a t^{2} \] - Part (d) $V$ What is the maximum height the ball reaches in meters? \[ h_{\text {mix }}=1000 \mathrm{~m} \] Part (e) Enter an expression for the elapsed time it takes for the ball to travel from the ground to a given height in terms of the initial velocity $v_{i}$, the velocity the ball has at that height $m_{\text {}}$, and the acceleration $a$. $\Delta t=$ $\square$
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Solution

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Solution Steps

Step 1: Determine the acceleration of the ball in the air

The acceleration of the ball in the air is due to gravity, which is \( g = -9.8 \, \text{m/s}^2 \). The negative sign indicates that the acceleration is directed downward.

Step 2: Determine the velocity of the ball at the highest point

At the highest point of its trajectory, the velocity of the ball is zero because it momentarily stops before starting to fall back down. Therefore, \( v_{\text{top}} = 0 \, \text{m/s} \).

Step 3: Derive the expression for the height of the ball as a function of time

The height \( h(t) \) of the ball as a function of time \( t \) can be expressed using the kinematic equation: \[ h(t) = v_i t + \frac{1}{2} a t^2 \] where \( v_i \) is the initial velocity and \( a \) is the acceleration.

Final Answer

  1. The acceleration of the ball in the air is \( -9.8 \, \text{m/s}^2 \).
  2. The velocity of the ball at the highest point is \( 0 \, \text{m/s} \).
  3. The height of the ball as a function of time is \( h(t) = v_i t + \frac{1}{2} a t^2 \).
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