Finding launch angle for ball bouncing 0.70 m away and reaching 3.10 m height

[nc1617]

Homework Statement

A ball is tossed so that it bounces off the ground, rises to a height of 3.10 m, and then hits the ground again 0.70 m away from the first bounce.

Homework Equations

H= .5(v^2/g) for y velocity.

The Attempt at a Solution

Vy=squareroot of(60.822) Vy= 7.799 m/s
Vx= .70 m/1.59 s= .441 m/s
inversetangent(Vy+Vx)= angle ------ is this correct?

i did the inversetangent and it came out to be about 79.4 degrees. but that wasn't the correct answer :(

 

[issacnewton]

your Vx , Vy are correct but the angle is wrong.

$$\tan(\theta)=\frac{V_y}{V_x}=\frac{7.79}{0.44}$$

 

[nc1617]

thank you! :)