A ball is kicked vertically 20 m/s from height of 1,5 m above the surface. Find a) max height the ball reaches, b) speed it hits the surface, c) time in the air

Question

How to solve these?

Answers: a) 21,9 m b) 20,7 m/s c) 4,1 s

Answer 1

Part 1 of the question:

You can solve this in two different ways.

a) Use Conservation of Energy Principle:

At the time the ball is kicked, it's energy its energy is a combination of potential (due to height above the ground) and Kinetic energies (due to the kick).

Initial Energy of the ball = mg.(1.5) + 1/2*m.(20)²

= 1.5mg + 200.m

At the peak height just before the ball begins to fall back, it's velocity is zero.The entire energy is stored as potential energy given by

Final Energy of the ball = mg.(h + 1.5)

Conservation of energy requires

Initial energy = final energy

1.5mg + 200.m = mg.(h+1.5)

With g= 9.8m/s², solve for 'h'

h = 20.41 m

Note that the maximum height that the ball achieves must include the initial height the ball is kicked from.

Max height achieved by ball = 20.41 + 1.5 = 21.91m

B). Use Newton's Equation of Motion

v² - u² =2as, where

final velocity, v = 0 (at the highest point),

initial velocity, u = 20m/s.

Acceleration 'a' = 9.8m/s² (same as gravity)

distance s = height reached above the point from where it is thrown up.

This yields identical results.

Part 2 of the question can also be approached similarly as above. In this case,

Initial Velocity = 0

Final Velocity = V

Height that the ball falls through = 21.91m (from previous answer)

Use v² - u² = 2gh to get

v = 20.72 m/s

Part 3 of the question requires you to calculate and add the times that the ball takes to reach it's highest point and then takes to fall back to ground. This can be done using Newton's equeation of motion

s = ut + 1/2*at²

When going up, s = 20.41m, u=20m/s, a = 9.8m/s²

When coming down, s = 21.91, u = 0, a = 9.8m/s²

You should be able to calculate and add times to answer part C.

Answer 2

a) Use equation v² = v_o² + 2a Δy, with v = 0 at top of flight: 0 = 400 + 2(-9.8) Δy, Δy = 400/19.6 = 20.4, so height above ground y = Δy +1.5 = 21.9 m.

b) With same equation and Δy = -1.5 from start to finish, v² = 20² -19.6*(-1.5) = 429.4, v = 20.7 m/s.

c) Use Δy = v_o t + 1/2 a t² = 20t + (-9.8/2) t² = 20t - 4.9t² = -1.5, then rearrange to quadratic format 4.9t² -20t - 1.5 = 0. Quadratic formula reduces to t = (20 ± 20.72)/9.8 and chose the larger answer t = 4.16 s (the small negative answer is when the ball left the ground if it had been kicked from there.