Question

In the figure here, a ball is thrown up onto a roof, landing 4.70 s later at height h = 25.0 m above the release level. The ball's path just before landing is angled at θ = 58.0˚ with the roof. (a) Find the horizontal distance d it travels. (Hint: One way is to reverse the motion, as if it is on a video.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's initial velocity?

Answer 1

(a) $y=y_0+v_0\sin\theta \cdot t-gt^2/2\to0=25+v_0\sin58°\cdot4.7-9.8\cdot4.7^2/2\to$

$v_0=20.88\ (m/s)$ .

$v_x=20.88\cdot\cos58°=11.07\ (m/s)$ . So, $d=11.07\cdot4.7=52\ (m)$

(b) $g=\Delta v_y/\Delta t\to -9.8=(v_{fy}-20.88\cdot\sin58°)/4.7\to v_{fy}=-28.36\ (m/s)$

$v=\sqrt{11.07^2+28.36^2}=30.44\ (m/s)$

(c) $\tan\alpha=v_y/v_x=28.36/11.07=2.56\to \alpha=68.7°$