Projectile Motion - motion on a horizontal surface

[bonbloc]

Projectile Motion - motion on a horizontal surface please help!

Homework Statement

Isaac Newton shot a dirty snowball 76m across a horizontal field. It passed directly over Edmund Haley's head, and was in the air for 7.6s. How high did it get above the field? Answer in m.

Homework Equations

Δd = V1(Δt) + 1/2(g)(Δt)^2
and i think some projectile motion equations, not sure if those will help.

The Attempt at a Solution

since at maximum height the velocity will be zero i tried this:
Δd = 0m/s(Δt) - 1/2(9.8m/s^2)(7.6s)^2
Δd = 283m
but this is not the right answer, so i would appreciate it if someone could help me with this.
Thank you.

 

[oli4]

Hi bonbloc
At maximum height, velocity (its vertical component, that is) is 0
By symmetry, you can easily see that the maximum height is also reached right in the middle of the trajectory.
So you will have v=0 at T/2 (7.6/2)
From this, you should be able to calculate the vertical component of the velocity.
Write then the equation of motion y(t) which will require this initial value and plug in the found velocity and T/2

Cheers...

 

[rude man]

Think of the snowball at its high point. It took 7.6/2 = 3.8s to get there (symmetrical trajectory). Now it's stationary & beginning to fall. How far does it fall in the remaining 3.8 sec?

It doesn't matter how far it went horizontally, or what its horizontal velocity was.