Questions: 3. A boy stands at the edge of a 25.0 m cliff and tosses a baseball straight upward with an initial speed of 22.0 m / s. Ignore the effects of air resistance. (a) How high will the ball go above the cliff edge? (b) What is the acceleration of the ball when it is at its maximum height? (c) Determine the position of the ball 4.70 s after it is thrown. Is the ball above or below the cliff edge at this time? (d) What is the velocity of the ball immediately before it impacts the ground at the base of the cliff?

Solution Steps

To find the maximum height the ball reaches above the cliff edge, we use the kinematic equation for vertical motion:

\[ v^2 = u^2 + 2a s \]

where:

• \( v = 0 \, \text{m/s} \) (final velocity at the maximum height),
• \( u = 22.0 \, \text{m/s} \) (initial velocity),
• \( a = -9.81 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it is directed downward),
• \( s \) is the displacement (height above the cliff edge).

Rearranging the equation to solve for \( s \):

\[ 0 = (22.0)^2 + 2(-9.81)s \]

\[ s = \frac{(22.0)^2}{2 \times 9.81} = \frac{484}{19.62} \approx 24.6735 \, \text{m} \]

The acceleration of the ball at any point in its trajectory, including at the maximum height, is due to gravity. Therefore, the acceleration is:

\[ a = -9.81 \, \text{m/s}^2 \]

To find the position of the ball after 4.70 seconds, we use the kinematic equation:

\[ s = ut + \frac{1}{2} a t^2 \]

where:

• \( u = 22.0 \, \text{m/s} \),
• \( a = -9.81 \, \text{m/s}^2 \),
• \( t = 4.70 \, \text{s} \).

Substituting the values:

\[ s = 22.0 \times 4.70 + \frac{1}{2} \times (-9.81) \times (4.70)^2 \]

\[ s = 103.4 - 108.26715 \approx -4.86715 \, \text{m} \]

Since the displacement is negative, the ball is below the cliff edge.

Final Answer