Mastering physics Ch 3 frustation

[Slizzardman]

Homework Statement

A quarterback passes the football downfield at 20 m/s. It leaves his hand 1.8 m above the ground and is caught by a receiver 30 m away at the same height. What is the maximum height of the ball on its way to the receiver?

Xi=0m
Xf=30m
Yi=1.8m
Yf=1.8m
Vi= 20 m/s
t=1.5s (time in the air, aka delta t)

Homework Equations

Yf=Yi+(Vyi)t-.5g(t^2)
xf=xi+(vx)i*t

The Attempt at a Solution

xf-vi=t=1.5s
since this was a parabolic motion I used .75s as the time for max height. I plugged the numbers in and go 4.55625, or 4.56m max height. This came from the original 1.8m height added to .5g*(.75s^2) from the first equation, as Yi=1.8 and Vyi atthe halfway point of .75s was 0m/s.

There were no angles given. How have I screwed this up? Apparently my answer is wrong.

 

[cepheid]

Welcome to PF!

You've used 20 m/s as the initial x-velocity in order to solve for the time in the air. However, 20 m/s is the total initial velocity, not just the velocity in the x-direction.

 

[cepheid]

There is an expression for the range of the projectile in terms of the initial speed and the launch angle. You don't know the launch angle, but you can solve for it, since you are given the range.