Question

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at 35 degrees below the horizon and the roof edge is 2.50 meters above the ground, find: (a) the time the baseball spends in the air, and (b) the horizontal distance from the roof edge to the point where the baseball lands on the ground.

Answer 1

Answer:

a) t = 0.528 s

b) D = 1.62 m

Explanation:

given,

speed of the baseball = 3.75 m/s

angle made with the horizontal = 35°

height of the roof edge = 2.5 m

using equation of motion

$s = ut +\dfrac{1}{2}gt^2$

$2.5 = vsin\theta \ t +\dfrac{1}{2}gt^2$

$2.5 = 3.75\ sin35^0 \ t +\dfrac{1}{2}\times 9.8 t^2$

4.9 t² + 2.15 t - 2.5 = 0

on solving the above equation

t = 0.528 s

b) D = v cos θ × t

D = 3.75 × cos 35° ×0.528

D = 1.62 m