Question

A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was it's intial velocity? How much total time will it take to come down? ​

Answer 1

Given: Distance travelled by the ball (s) =

4.05 m.

acceleration a = g 10 m/s²

To find:

1. Initial velocity (u)

2. Time taken (t)

Formulae:

1. v² = u² + 2 as

2.

$s = ut \: + \frac{1}{2} {at}^{2}$

Calculation: For upward motion of the ball.

(v) = 0.

a = -g = -10 m/s2

From formula (i),

0=u²+ 2(-10) x 4.05

:. 1² = = 81

= 9 m/s

For downward motion of the ball, (u) = 0.

a = g = 10 m/s²

From formula (ii).

$4 .05 \: = 0 \times \frac{1}{2} \times 10t {}^{2}$

$t { }^{2} = \frac{4.05}{5} = 0.81$

t = 0.9 s

Time for upward journey of the ball will be the same as time for downward journey

i. e., 0.9 s.

Total time taken = 2 x 0.9 = 1.8 s

1. The initial velocity of the ball is 9 m/s.

2. The total time taken by the ball to reach

the ground is 1.8 s.

Answer 2

Given :-

For the upward motion of the ball,the final velocity of the ball = v = 0

Distance travelled by the the ball = 4.05 m

acceleration a = g = -10 m/s²

Using Newton's first equation of equation of motion

$\\ \dashrightarrow \:\bf \red{v² = u² + 2 a s }$

$\\ \dashrightarrow \:\bf \red{v² = u² + 2 a s0 = u² +2 (-10) × 4.05}$

Therefore u² = 81

u = 9m/s the initial velocity of the ball is 9m/s

Now let us consider the downward motion of the ball.Suppose the ball takes t seconds to come down. Now the initial velocity of the ball is zero, u = 0. Distance traveled by the ball reaching the ground = 4.05 m. As velocity and acceleration are in same direction,

a = g = 10m/s

According to the Newton's second law of motion

${{{{{\red{S=ut \: \dfrac{1}{2} \: at²}}}}}}$

$\begin{gathered}\end{gathered}$

${{{{\sf{\red{4.05=0 + \: \dfrac{1}{2} \: 10t {}^{2} }}}}}}$

$\begin{gathered}\end{gathered}$

${{{{\sf{\red{t {}^{2} = \: \dfrac{4.05}{5} \: = 0.81 \: , \: \: t = 0.9s}}}}}}$

$\begin{gathered}\end{gathered}$

The ball will take 0.9s to reach the ground. it will take the same time to go up. Thus, the total time taken = 2 × 0.9 = 1.8 s