Question

A stone is thrown from the top of a tower at an angle of ${30}^0$ up with the horizontal with a velocity of $16m/s$. After $4$ seconds of flight it strikes the ground. Calculate the height of the tower from the ground and the horizontal range of the stone $(g=9.8m/s^2)$

Answer 1

Given $V=16m/s$, time $t=4s$. Let height $h$, and Horizontal range $R$

Velocity in upward direction $V_y=Vsin\left({30}^o\right)=8m/s$,

Velocity in horizontal $V_x=Vcos\left({30}^o\right)=8\sqrt{3}$

In upward direction, Using $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow h=8\times 4-\frac{1}{2}\times 9.8\times 4^2$ $\Rightarrow h=-46.4$ , Hence, Height$=46.4m$,

In horizontal direction, $Range=V_x\times t$

$\Rightarrow R=8\sqrt{3}\times 4=32\sqrt{3}m$, Hence Range is $32\sqrt{3}m$