Question

Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 34.6 m/s2 for a height 61.0 cm . He releases it at height 2.30 m above the ground. You may ignore air resistance.

A) What is the speed of the shot when Sam releases it?
B) How high above the ground does it go?
C) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

Answer 1

6.49707626552 m/s

4.45147808359 m

1.46212197842 s

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 34.6* 0.61+0^2)\\\Rightarrow v=6.49707626552\ m/s`

The speed with which Sam releases the ball is 6.49707626552 m/s

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2g)\\\Rightarrow s=(0^2-6.49707626552^2)/(2* -9.81)\\\Rightarrow s=2.15147808359\ m`

The ball reaches a height of 2.3+2.15147808359 = 4.45147808359 m

Distance to his head from the maximum height is 4.45147808359-1.83 = 2.62147808359 m

`s=ut+(1)/(2)gt^2\\\Rightarrow 2.62147808359=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(2.62147808359* 2)/(9.81)}\\\Rightarrow t=0.731060989212\ s`

Time taken to go down to his head is 0.731060989212 seconds from the maximum height

Time taken from the moment the ball left his head height = 0.731060989212+0.731060989212 = 1.46212197842 s