<p>ball is thrown up onto a roof, landing 3.10 s later at height h = 19.0 m above the release level. The ball's path just before landing is angled at θ = 61.0° with the roof. (a) Find the horizontal distance d it travels. (Hint: One way is to reverse the motion, as if on videotape.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's initial velocity?</p>
<p>thanks in advance.</p>
<p>The horizontal component of velocity can’t change since no acceleration in horizontal thus,
theta would be 61 degrees.</p>
<p>If the initial velocity was V, then for motion in horizontal </p>
<p>V cos(61) t = d</p>
<p>For motion in vertical,</p>
<p>h = V sin(61) t - 0.5 g t^2</p>
<p>From these two, eliminate V first to get</p>
<p>d = cot(61) [ h + 0.5 g t^2] all of h , g and t are known.</p>
<p>Use the first equation,
V cos(61) t = d after you have found out d above to get the initial velocity V.</p>