Question

A 3 kg cannonball is shot straight up from ground level at a
velocity of 30 m/s. For the following questions, neglect air
resistance and assume acceleration due to gravity is -10
m/s2.

(a) Use the 1D equations of motion to find:

i. The maximum height the cannonball reaches

ii. The time it takes to reach this height

iii. The time it takes to fall back to the ground

iv. The speed with which it hits the ground

Answer #1

i)

V_{i} = initial velocity = 30 m/s

a = acceleration due to gravity = -10 m/s^{2}

V_{f} = final velocity at the maximum height = 0 m/s

H = maximum height gained

Using the equation

V_{f}^{2} = V_{i}^{2} + 2 aH

(0)^{2} = (30)^{2} + 2 (- 10) H

H = 45 m/s

ii)

t = time taken to reach maximum height

using the equation

V_{f} = V_{i} + a t

0 = 30 + (-10) t

t = 3 sec

iii)

consider the motion from Top to bottom :

V_{i} = initial velocity = 0 m/s

a = acceleration = - 10

Y = displacement = - 45

t = time taken

using the equation

Y = V_{i} t + (0.5) a t^{2}

- 45 = 0 t + (0.5) (- 10) t^{2}

t = 3 sec

iv)

V_{f} = final velocity as it hits the ground

using the equation

V_{f} = V_{i} + at

V_{f} = 0 + (- 10) (3)

V_{f} = - 30 m/s

speed = 30 m/s

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