A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entire time of motion? What is the velocity at 5 seconds after the projection? Take g = 10m/s²

A body is projected vertically up.

∴ According to the question;

The initial speed (u) = 40 m/s

Acceleration due to gravity (g) = 10 m/s² .

From Newton’s third equation of motion;

V² = u² + 2as;

In our question;

V = 0 (Velocity at maximum height is zero);

U = 40 m/s.

A = -g = -10 m/s². (When object will be going up the acceleration due to gravity will be acting downwards to make an object to fall. Hence by sign convention direction of motion and acceleration is opposite therefore a is negative)

S = Maximum Height (H).

Putting the values in the equation we get.

0² = 40² – 2 x 10 x H.

⇒ 0 = 1600 – 20H.

⇒ 1600 = 20H.

⇒ H = .

Maximum height reached by the body is 80m.

Total time (T ) =

⇒ T =

⇒ T =

The entire time of motion is 8 s

Time is taken to reach maximum height =

=

Thus it takes 4 sec to reach the maximum height and returns with 0 initial speed.

∴ speed after 1 sec in return = speed after 5 seconds of throwing the ball.

⇒ speed (v) = u + at;

∵ u = 0 and t = 1sec; a = 10 m/s² .

⇒ v = 0 + 10 x 1

⇒ v = 10 ms.

Speed after 5 sec of throwing of the ball is 10 m/s.