Question

A ball of mass 50 gram is thrown vertically upwards with an initial velocity of 20 m/s.

Calculate i) the initial kinetic energy imparted to the ball ii) the maximum height reached

if a friction is neglected.​

Answer 1

Answer:

Given that,

m=0.05 and v=20m/s

1

Maximum potential energy at highest point= Initial kinetic energy at the bottom point

∴Maximum potential energy =

2

1

mv

2

=

2

1

×0.5×20×20

=

2

200

=100J

December 26, 2019

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Answer 2

Question :

A ball of mass 50 g is thrown vertically upwards with an initial velocity of 20 m/s. Calculate (i) the initial kinetic energy imparted to the ball, (ii) the maximum height reached if air friction is neglected and (iii) the maximum height reached if 40% of the initial energy is lost against the air friction.

Solution :

Mass (m) = 50 g = 50/1000 Kg = 0.05 Kg

Initial velocity (u) = 20 m/s

Acc. due to gravity (g) = 10 m/s²

(i) $\frac{1}{2}mu^2$ = $\frac{1}{2}\times0.05\times20^2$

• 10 J

(ii) If air friction is negligible, then potential energy at the maximum height is equal to initial kinetic energy.

$mgh=\frac{1}{2}mv^2$

$h=\frac{v^2}{2g}$ $=\frac{20^2}{2\times10}$

• 20 m

(iii) If 40% of the initial energy is lost against the air friction, potential energy at maximum height is equal to 60% of the initial kinetic energy.

$mgh=\frac{60}{100}\times\frac{1}{2}mv^2$

$h=\frac{0.6v^2}{2g}$$=\frac{0.6\times20^2}{2\times10}$

• 12 m