Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m 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?

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.