Question

A ball is thrown in vertically upwards direction with velocity 50 m/s. Maximum height covered

by the ball will be ………… (g = 10 m/s2

)​

Answer 1

Explanation:

(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,

final velocity at the highest point (v)=0

So applying the 3rd equation of motion we get:

v

2

=u

2

−2gh

max

⇒0=50

2

−2×10×h

max

⇒h

max

=

20

2500

=125m

(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:

v=u−gt

⇒0=50−10t

⇒t=5s

(c) Let speed at half of max height be V then:

V

2

=50

2

−2g

2

125

⇒V

2

=2500−1250=1250

⇒V=

1250

=35.35m/s.

Answer 2

Answer:

A ball is thrown in a vertically upwards direction with a velocity of 50 m/s. Maximum height covered  by the ball will be $125\ m$

Explanation:

• According to the third law of motion,

        $v^{2} - u^{2} =2as$    ...(1)

• where v is the final velocity of the object, u is the initial velocity of the object, a is the acceleration and s is the distance covered by the object.

In the given problem,

$s = h_{max}$  

$a =- g$ (since thrown upward)

where g is the acceleration due to gravity and $h_{max}$ is the maximum height covered by the ball.

Given,

$v = 0$

$u = 50\ m/s$

$g = 10\ m/s^2$

Substitute these values into equation(1)

$0^{2} - 50^{2} =2\times-10\times h_{max}$

$h_{max}= \frac{50^{2} }{20} = 125\ m$