Question

problem #14

• a football is thrown with an initial upward velocity component of 15.0 m/s and a horizontal velocity component of 18.0 m/s (a) how much time is required for the football to reach the highest point in its trajectory? (b) how high does it get above its release point? (c) how much time after it is thrown does it take to return to its original height? how does this time compare with what you calculated in part (b)? is your answer reasonable? (d) how far has the football traveled horizontally from its original position?

Explanation:

(a) Step1: Vertical velocity at max height is 0

\( v_y = v_{0y} - gt = 0 \)

(a) Step2: Solve for time to max height

\( t = \frac{v_{0y}}{g} = \frac{15.0}{9.8} \approx 1.53 \, \text{s} \)

(b) Step1: Use vertical velocity-displacement equation

\( v_y^2 = v_{0y}^2 - 2gy \), \( v_y = 0 \)

(b) Step2: Calculate max height

\( y = \frac{v_{0y}^2}{2g} = \frac{15.0^2}{2 \times 9.8} \approx 11.5 \, \text{m} \)

(c) Step1: Total time to return is 2×up time

\( t_{\text{total}} = 2t \) (t from (a))

(c) Step2: Compute total time and compare

\( t_{\text{total}} = 2 \times 1.53 \approx 3.06 \, \text{s} \); Equal to 2×(a) time, reasonable.

(d) Step1: Horizontal distance = horizontal velocity×total time

\( x = v_{0x} \times t_{\text{total}} \)

(d) Step2: Calculate horizontal distance

\( x = 18.0 \times 3.06 \approx 55.1 \, \text{m} \)

Answer:

(a) 1.53 s
(b) 11.5 m
(c) 3.06 s; It is twice the time calculated in part (a), which is reasonable.
(d) 55.1 m