Question

A mortar shell is accidentally shot straight up with an initial speed of 147 m/s. How long does it take for the shell to reach its maximum height? How high does the shell rise? If the shell falls straight down after is stops, what is its velocity when it hits the ground? What is the total time the shell is in the air?

Answer 1

1) t ≈ 14.985 seconds

2) The maximum height reached ≈ 1101.4 meters

3) The velocity with which it hits the ground = 147 m/s

4) The total time in the air ≈ 29.969 seconds

Step-by-step explanation:

1) The given information are;

The initial velocity of the mortar shell = 147 m/s

The direction of the mortar shell = Vertically upwards

Therefore, we have, from the equation of motion

v = u - g·t

Where;

v = The final velocity = 0 at maximum height

u = The initial velocity = 147 m/s

g = The acceleration due to gravity = 9.81 m/s²

Therefore, we have;

0 = 147 - 9.81 × t

9.81 × t = 147

t = 147/9.81 ≈ 14.985 seconds

t ≈ 14.985 seconds

2) From the equation of motion, v² = u² - 2×g×s

Where;

s = The maximum height reached

We have;

0² = 147² - 2×9.81×s

2×9.81×s = 147²

s = 147²/(2×9.81) ≈ 1101.4 meters

The maximum height reached ≈ 1101.4 meters

3) The velocity with which it hits the ground is equal to the initial velocity = 147 m/s

4) The total time in the air = 2 × The time it takes to reach maximum height

The total time in the air = 2 × 14.985 ≈ 29.969 seconds

The total time in the air ≈ 29.969 seconds