Question

from the edge of the rooftop of a building, a boy thros a stone at an angle of 25 above the horizontal. The stone hits the ground 4.2s later, 105m away from the base of the builing. Find the final speed of the stone

Answer 1

Answer:38.675 m/s

Explanation:

$\theta =25^{\circ}$

t=4.2 s

$R_x=105m$

Solving in x direction

$R_x=ut+\frac{1}{2}\left ( 0\right )t^2$

$105=ucos25\times 4.2$

ucos25=25

u=27.584 m/s

Initial velocity in vertical direction usin25

Let h be the height of Rooftop

$h=\left ( -usin25\right )4.2+\frac{1}{2} \left ( 9.8\right )\left ( 4.2\right )^2$

h=37.47 m

Therefore final vertical velocity is $v_{vf}^2-\left ( usin25\right )^2=2\times 9.81\times 37.47$

$v_{vf}$=29.51 m/s

Final Resultant velocity

$V_{final}=\sqrt{29.51^2+24.99^2}$

$V_{final}=38.675 m/s$