Question

part A,B and C please

Part A Constants What is the speed of the shot when he releases it? Sam heaves a shot with weight 16-lb straight upward, giving ita constant upward acceleration from rest of 42.8 m/s for a height 68.0 cm. He releases it at height 2.25 m above the ground. You may ignore air resistance. u= m/s Submit Part B How high above the ground does it go? Submit PartC How much time does he have to get out of its way before it retums to the height of the top of his head, a distance 1.80 m above the ground?

Answer 1

Upward acceleration, a = 42.8 m/s^2

Distance through which acceleration is given, d = 68.0 cm = 0.68 m

Part A -

Speed on release -

Vf^2 = Vi^2 + 2* a* d

=> Vf^2 = 0 + 2(42.8)(0.68) = 58.208

=> Vf = 7.63 m/s

Part B -

how high -

Vf^2 = Vi^2 + 2 g d

Vi is the Vf from above

0 = 58.208 + 2(-9.81) d

=> 19.62 d = 58.208

=> d = 2.97 m
h = d + release height = 2.97 m + 2.25 m = 5.22 total

Part C -

Time to fall back to head height -

Distance from top to head high = 5.22 m - 1.80 m = 3.42 m
the time he needs to get out of the way is time to top + time to fall back to head high
time to top
g = deltaV / t
t = de;taV / g = 7.63 / 9.81 = 0.78 s
time to fall back to head high
d = 1/2 g t^2
3.42 = 1/2(9.81) t^2
t^2 = 0.697 s
t = 0.835 s
total t = 0.78 s + 0.835 s = 1.615 s