Question

A rock is tossed straight up from the ground with a speed of 11 m/s . When it returns, it falls into a hole 10 m deep. A rock is tossed straight up from the ground with a speed of 11 m/s . When it returns, it falls into a hole 10 m deep.A rock is tossed straight up from the ground with a speed of 11 m/s . When it returns, it falls into a hole 10 m deep. What is the rock's velocity as it hits the bottom of the hole?

Answer 1

Answer:

17.8101094888 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-11^2}{2\times -9.81}\\\Rightarrow s=6.16717635066\ m$

Total distance of the fall is 6.16717635066+10 = 16.16717635066 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.16717635066=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.16717635066\times 2}{9.81}}\\\Rightarrow t=1.81550555441\ s$

Time taken to fall total distance 1.81550555441 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.81550555441\\\Rightarrow v=17.8101094888\ m/s$

The velocity of the rock when it hits the bottom of the hole is 17.8101094888 m/s