Question

I can’t figure this out!!! Answer what you can , please.

Answer 1

a/b. The ball has velocity vector at time $t$

$\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)$

where $v_0=16\dfrac{\rm m}{\rm s}$ is the ball's initial speed and $g=9.8\dfrac{\rm m}{\mathrm s^2}$.

c. At its highest point, the ball has 0 vertical speed. This occurs when

$v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s$

d. Recall that

${v_y}^2-{v_{0y}}^2=-2g\Delta y$

so that at its highest point,

$0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m$

e. This is just twice the time it takes for the ball to reach its maximum height, $t=2.9\,\mathrm s$.

f. The ball's horizontal position after time $t$ is

$v_0\cos63^\circ\,t$

so that after the time found in part (f), the ball has traveled

$v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m$