~~~~~physics problem

An object is projected vertically upwards with a speed of 24.5m s^-1 near a 29.4m tall tower. calculate the time during which the object is above the tower( ?S interval between t=? and t=?)

Let the velocity at the top of the tower is v m/s
=>By v^2 = u^2 - 2gh
=>v^2 = (24.5)^2 - 2 x 9.8 x 29.4
=>v = √24.01
=>v = 4.9 m/s
=>Let the time to reach the highest point from the top is t sec
=>By v = u - gt
=>0 = 4.9 - 9.8 x t
=>t = 0.5 sec
=>the time during which the object is above the tower = 2t = 2 x 0.5 = 1 sec

Well, we start by finding the velocity at which the ball is travelling when it reaches the height of the tower;

S = 29.4

S = ut + 0.5at^2, so the time t can be found as;

24.5t - (4.9 * t^2) = 29.5

4.9t^2 - 24.5t + 29.5 = 0

This is a standard quadratic formula thing, so we can find that t = 2.02 seconds.

This means that v at 2.02 seconds = u + at = 24.5 - (2.02 * 9.8) = 24.5 - ~20 = 4.5

So now we know that the time spent above the tower until the apex is

4.5 / a = 0.46

So double this is the time spent above the tower, which is 0.92 seconds.



Also;

Knowing a, t, and u we can now again use;

S = ut + (0.5a t^2)

So S = 0.46 * 4.5 - (4.9 * 0.46^2)

= 2.07 - 1.04 = 1.03

So, 1.03m above tower.