A stone of mass 1.0kg is dropped from a certain height and takes 8 seconds for the stone to strike the sandy ground, upon reaching the ground the stone penetrate 5.0 into the ground ( take g=10m/s) A)calculate de height from where the stone is dropped?

Question

A stone of mass 1.0kg is dropped from a certain height and takes 8 seconds for the stone to strike the sandy ground, upon reaching the ground the stone penetrate 5.0 into the ground ( take g=10m/s) A)calculate de height from where the stone is dropped?

1 Answer

Impact of this question

3188 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T22:15:13

$H_{total} = 325 m$ $H_"total"=325m$

Explanation:

Acceleration due to gravity $= + g = 10 m / s^{2}$ $=+g=10m//s^2$
Initial velocity $= 0 m / s$ $=0m//s$
Time taken for Stone to hit the ground $= 8 sec s$ $=8secs$
$H_{1} = 5 . o m$ $H_1=5.om$
Applying eqns of motion under gravity
V $=$ $=$ u + gt
V $=$ $=$ $0 +$ $0 +$ ( $10$ $10$ x $8$ $8$ )
V $=$ $=$ $80 m / s$ $80m//s$
Recall:
$v^{2} = u^{2} + 2 g H$ $v^2=u^2 +2gH$
Let $H_{2} =$ $H_2=$ height to ground level
$80^{2} = 0^{2} + 2$ $80^2=0^2 + 2$ ( $10 H_{2}$ $10H_2$ )
$6400 - 0 = + 20 H_{2}$ $6400-0= + 20H_2$
$6400 = 20 H_{2}$ $6400=20H_2$
$H_{2} = \frac{6400}{20}$ $H_2=(6400)/20$
$H_{2} = 320 m$ $H_2=320m$
Total height $= H_{1} + H_{2}$ $=H_1+H_2$
$= (320 + 5) m$ $=(320+5)m$
$= 325 m$ $=325m$

Science

Math

Humanities

... and beyond

A stone of mass 1.0kg is dropped from a certain height and takes 8 seconds for the stone to strike the sandy ground, upon reaching the ground the stone penetrate 5.0 into the ground ( take g=10m/s) A)calculate de height from where the stone is dropped?

1 Answer

Explanation:

Related questions

Impact of this question