Question

A girl, standing on the roof of a house which is 5.00 m high, throws a stone vertic upward. It strikes the ground 7.00 s after being thrown. Calculate: The initial velocity of the stone The maximum height the stone reaches above a. b. i. The roof ii. The ground and c.

Answer 1

Answer:

Explanation:

Given

Height of roof=5 m

Time taken by stone to reach ground 7 s

Let u be the initial velocity

therefore maximum height is h

$v^2-u^2=2as$

$0-u^2=2(-g)h$

$u=\sqrt{2gh}$

time taken to reach max height

v=u+at

0=u-gt

$t=\frac{u}{g}$

Now time taken to reach ground is $t_2$

$h+5=0\times t_2+\frac{gt_2^2}{2}$

$\frac{u^2}{2g}+5=\frac{gt_2^2}{2}$

$t_2=\sqrt{\frac{u^2+10g}{g^2}}$

$t_2=\frac{\sqrt{u^2+10g}}{g}$

total time is $t+t_2=7$

$\frac{u}{g}+\frac{\sqrt{u^2+10g}}{g}=7$

$\frac{\sqrt{u^2+10g}}{g}=7-\frac{u}{g}$

Squaring both side

$u^2+10g=49g^2+u^2-14ug$

$14u=49g-10$

u=33.58 m/s

therefore $h=\frac{u^2}{2g}$

h=57.55 m

maximum height from ground is 57.55+5=62.55 m