Answer to Question #156856 in Mechanics | Relativity for Gill Allain

Question

A golf player hits a ball from a point O on a horizontal ground with a velocity of v(13)^1/2 at an angle u above the horizontal where tanu = 3/2. The ball first hit tge ground at a point A, where OA = 120m.
(a) Show that its height h above the ground at time t seconds is given by h = (30t - 5t^2)m. Hence or otherwise find:
(b) the maximum height reached by the ball above the level of projection.
(c) the time of flight.

Answer 1

Given. $v_0=36\ (m/s)$ ; $\tan u=3/2\to u=56.3°$ ; $s=120(m)$

a) $h=v_{0y}t-gt^2/2=v_0\cdot\sin u\cdot t-gt^2/2=$

$=36\cdot\sin56.3°\cdot t-9.8\cdot t^2/2\approx30t-5t^2$ . Answer

b) $h_{max}=\frac{v_{0y}^2}{2g}=\frac{(v_0\cdot\sin u)^2}{2g}=\frac{(36\cdot\sin 56.3°)^2}{2\cdot 9.8}=45.8(m)$ .Answer

c) $t=120/(36\cdot\cos56.3°)\approx6(s)$ . Answer