Question

With what speed should a body be thrown upwards so that the distances traversed in 5th and 6th second are equal ?

Answer 1

The body must be thrown with a speed of 50 metres per second.

 

The above problem makes use of the laws of kinematics and Newton’s laws of motion to calculate various parameters involved in vertical projectile motion.

 

Examining the motion of body, we see that it travels the same distance in the 5th and 6th seconds. Hence, the body should reach its maximum height by the end of the 5th second. Hence, the upward motion of the body has its time of flight as 5 seconds while the downward motion also has 5 seconds as the time of flight.

 

We must calculate the speed with which the body should be thrown so that it comes to rest in 5 seconds while acting against a deceleration of 10 mps^2.

 

Final speed (v) = 0 mps

Initial speed (u) = u mps

Deceleration (a) = 10 mps^2

Time taken (t) = 5s

 

Using Newton’s First Law of Kinematics,

 

v = u - at

0 = u - 10 x 5

u = 50 mps