A ball is thrown upwards from the ground with an initial velocity ‘u’.The ball is at a height of 60 m at two times,the interval being 6 s.Find the initial velocity ‘u’

Solution:

First Method:

The ball reaches the given height of 60 m at a time t ₁ seconds ( rising up).For the second time it reaches the 60 m height at a time t ₂ seconds. (comes down)

Applying the equation of motion with the sign convention.

Initial velocity (u) = +u

Acceleration due to gravity (a) =-g

Displacement (s) =+60 m

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

$$60=ut+\frac{1}{2}(-10)t^2$$

$$60=ut-5t^2$$

$$5t^2-ut+60=0$$

We can apply the quadratic equation to find the roots of the above equation.

$$t={-b\pm\sqrt{b^2-4ac}\over 2a}$$

$$t={u\pm\sqrt{u^2-4(5)(600}\over 10}$$

$$t={u\pm\sqrt{u^2-1200}\over 10}$$

The above equation has two roots, which gives the two values of time t at a height of 60 m

The time interval as per question is 6 s. Subtracting the two values to get 6.

$${u+\sqrt{u^2-1200}\over 10}-{u-\sqrt {u^2-1200}\over 10}=6$$

$${\frac {u}{10}}+{\sqrt{u^2-1200}\over 10}-{\frac{u}{10}}+{\sqrt{u^2-1200}\over 10}=6$$

$$2{\sqrt {u^2-1200}\over 10}=6$$

$${\sqrt{u^2-1200}}=30$$

Squaring on both sides

$${u^2-1200}=900$$

$$u={\sqrt 2100}$$

Hence the ball is thrown upwards with an initial velocity of u=45.83 ms^-1

SECOND METHOD:

Distance travelled:

Distance travelled by the ball during the 3 seconds (Dropped case)

$${h=\frac{1}{2}gt^2}$$

$${h=\frac{1}{2}(10)3^2}$$

$${h=45 \ m }$$ (refer the top image)

Total height:

Total height in which the the ball is thrown with a velocity u is

$${60+45=105 \ m }$$

Calculating the initial velocity(u): (Thrown up case)

$${u=\sqrt{2gh}}$$

Squaring on both sides

$${u^2=2gh}$$

$${u=\sqrt{2gh}}$$

$${u=\sqrt{2(10)(105)}}$$

$${u=\sqrt{2100}}$$

$${u=45.83 \ \frac{m}{s}}$$

Hence the ball is thrown upwards with an initial velocity of u=45.83 ms^-1