Projectiles Problem: Time to Hit Ground from 45m

[omartharwat]

a ball is projected vitically from the top of a building with height of 45 m from Erath`s surface . it velocity is determind by the relation v= 40-10t how much time the ball will take to hit the ground?

 

[topsquark]

Well, we need an equation relating the velocity with the distance traveled. We can integrate:
[math]\int_0^t v(t) ~ dt = \int_0^t (40 - t) ~ dt \implies y(t) - y(0) = 40 t - 5t^2[/math]

If you can't use Calculus then consider that the acceleration ([math]a = \Delta v / \Delta t[/math]) is constant. So we can use y(t) - y(0) = v_0 t + 1/2 a t^2[/math], which gives the same thing.

Give it a try and show us what you get.

-Dan

 

[HallsofIvy]

There is a constant acceleration, $-g= -9.8 m/s^2$ so after time t, the speed is $v(t)= v_0- 9.8t m/s$ where $v_0$ is the initial speed. The problem says that the velocity is $40- 10t$. I suspect that "10" is an approximation to g= 9.8 so that $v_0= 40 m/s$.

The height after t seconds will be $h(t)= 45+ 40t- 5t^2$ (the "45" is the initial height at t= 0).

The "ground" is at h= 0 so to find the time until the ball hits the ground you need to solve the equation $h(t)= 45+ 40t- 5t^2= 0$ for t.

First divide both sides by -5 to get $t^2- 8t- 9= (t- 9)(t+ 1)= 0$. The roots are t= 9 and t= -1. Since the ball cannot hit the ground before it was thrown up, t= 9. The ball hits the ground 9 seconds after being thrown upward.

(t= -1 can be interpreted as when the ball could have been thrown up from the ground to be at 45 feet at t= 0.)