Question

Astone is thrown directly upward with an initial speed of 9.6 m/s from a height of 12.8 m. After what time interval (in s) does the stone strike the ground? Use g 9.8 m/s^2 Enter a number with 2 digits behind the decimal point.

Answer 1

1.89 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity = 9.6 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow 0=9.6-9.8* t\\\Rightarrow (-9.6)/(-9.8)=t\\\Rightarrow t=0.97 \s`

Time taken to reach maximum height is 0.97 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s=9.6* 0.97+(1)/(2)* -9.8* 0.97^2\\\Rightarrow s=4.7\ m`

So, the stone would travel 4.7 m up

So, total height ball would fall is 4.7+12.8 = 17.5 m

`s=ut+(1)/(2)at^2\\\Rightarrow 17.5=0t+(1)/(2)* 9.8* t^2\\\Rightarrow t=\sqrt{(17.5* 2)/(9.8)}\\\Rightarrow t=1.89\ s`

Time taken by the stone to travel 17.5 m is 1.89 seconds