Question

A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. The point of release is 45 m above the ground. How long does it take for the stone to hit the ground? What is the stone's speed?

Answer 1

Answer:

Explanation:

Given

inclination $\theta =30^{\circ}$

initial speed $u=20\ m/s$

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

$t_1=\frac{2u\sin \theta }{g}$

$t_1=\frac{2\times 20\times \sin 30}{10}$

$t_1=2\ s$

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

$h=u_yt+\frac{1}{2}a_yt^2$

where, $h=height$

$u_y=vertical velocity$

$a_y=vertical acceleration$

$t_0=time$

$45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2$

$t_0^2=\frac{70}{9.8}$

$t_0=2.64\ s$

thus total time time required is $t=t_0+t_1=2.64+2=4.64\ s$

vertical velocity just before hitting

$v_y=\sqrt{u_y^2+2\times a_y\times s}$

$v_y=\sqrt{10^2+2\times 10\times 45}$

$v_y=\sqrt{1000}=31.622\ m/s$

Horizontal velocity $v_x=u\cos 30=17.32\ m/s$

Net velocity Just before hitting $=\sqrt{v_x^2+v_y^2}$

$=\sqrt{(17.32)^2+(31.62)^2}$

$=\sqrt{1299.82}=36.05\ m/s$