Physics 1 Marks Question

1. Downward motion of Ball

Ball is dropped from a height, s = 90m

Initial velocity of the ball, u = 0

Acceleration, a = g = 9.8m/s²

Final velocity of the ball = v

From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:

s = ut + (1/2)at²

90 = 0 + (1/2) × 9.8t²

$\text{t} = \sqrt{18.38} = 4.29\text{s}$

From first equation of motion, final velocity is given as:

v = u + at

= 0 + 9.8 × 4.29 = 42.04m/s

Here we see that v=at, so velocity varies linearly with downward motion of ball.

This part of motion is represented by first line (1) in the below graph

2. First rebound

Now Rebound velocity of the ball, u_r =v(1 - 1/10) = 9v/10 = 9 × 42.04/10 = 37.84 m/s

This is represented the line (2) in the graph. We are assumed negligible time of collision between ball and floor

3. Upward motion of ball

Time (t₁) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

v = u + at₁

0 = 37.84 + (– 9.8)t₁

t₁ = -37.84/-9.8 = 3.86s

Total time taken by the ball = t + t₁ = 4.29 + 3.86 = 8.15s

This is represented by line (3) in the graph

4. Second descent of ball

As the time of ascent is equal to the time of descent, the ball takes 3.86s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor = 9 × 37.84/10 = 34.05m/s

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01s

This is represented by line (4) in the graph

3. Velocity is zero at t = 1 second

Explanation:

$\text{x}=3\text{t}^2-6\text{t},$ So (Velocity)_x $=\frac{\text{dx}}{\text{dt}}=6\text{t}-6$

(Acceleration)_x $=\frac{\text{d}^2\text{y}}{\text{dt}^2}=6\text{t};\text{y}=\text{t}^2-2\text{t};$

so (Velocity)_y $=\frac{\text{d}^2\text{y}}{\text{dt}^2}=2;$ At time t = 1,

$\frac{\text{dx}}{\text{dt}}=6\times1-6=0$ and $\frac{\text{dy}}{\text{dt}}=2\times1-2=0$