Question

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.30 m/s. If the roof is pitched at 36.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. 

a) Find the time the baseball spends in the air (in s) 

b) Find the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m)  

Answer 1

a) The horizontal component of the velocity of the ball at the roof edge= \(v_{0} \cos \theta=4.3 \cos 36^{0}=4.3 * 0.8=3.4 \mathrm{~m} / \mathrm{s}\).

The vertical component of the velocity \(=v_{0} \sin \theta=4.3 \sin 36^{0}=4.3 * 0.6=2.6 \mathrm{~m} / \mathrm{s}\).

The kinematical equation to be applied is \(y=y_{0}+u t+1 / 2 a t^{2}\) where \(y\) is the vertical distance travelled by the ball and \(u\) is the initial vertical velocity and a is the vertical acceleration of the ball which is equal to \(\mathrm{g}\).

Choosing the downward direction as positive and the origin of the coordinate system as the roof edge we obtain:

\(2.9=2.6 t+1.2 * 9.8 * t^{2}=2.6 t+4.9 t^{2}\) since \(y_{0}=0\) and \(u=v_{0} \sin 36°=2.6 \mathrm{~m} / \mathrm{s} .\)

Therefore we have the quadratic equation \(4.9 t^{2}+2.6 t-2.9=0 .\) On solving this equation we obtain \(t=0.6 \mathrm{~s}\).

This is the time the baseball spends in the air.

b) The horizontal distance travelled by the ball \(=v_{0} \cos 36° * t=3.4 * 0.6=2 \mathrm{~m}\).

It is to be noted that the horizontal and vertical motions of the baseball are independent of each other.