Question

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?

Answer 1

a) 0.658 seconds

b) 0.96 inches

Step-by-step explanation:

`v=u+at\\\Rightarrow 0=4.5-32.1* t\\\Rightarrow (-4.5)/(-32.1)=t\\\Rightarrow t=0.14 \s`

Time taken by the ball to reach the highest point is 0.14 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s=4.5* 0.14+(1)/(2)* -32.1* 0.14^2\\\Rightarrow s=0.315\ ft`

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

`s=ut+(1)/(2)at^2\\\Rightarrow 4.315=0t+(1)/(2)* 32.1* t^2\\\Rightarrow t=\sqrt{(4.315* 2)/(32.1)}\\\Rightarrow t=0.518\ s`

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

`v=u+at\\\Rightarrow v=0+32.1* 0.518\\\Rightarrow v=16.62\ ft/s`

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-16.62^2)/(2* -100* 3.28)\\\Rightarrow s=0.42\ ft`

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches