Help for a Physics Problem!?

Question

Please help me out! I am so confused :(

A ball is thrown straight upward and returns to the thrower's hand after 2.40 s in the air. A second ball is thrown at an angle of 44.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically (in m/s)?

Answer 1

OK, there are 2 portions to this question: find the speed of the first throw, then use trigonometry to solve for the speed of the second throw.

Step 1: Find speed of the first throw.
Using our velocity equation, v1 = v2t + 1/2 at^2, we can set the following:
a = -9.8m/s^2
t = 2.40s
v1 = -v2 (because it's going the opposite direction)

So:
v1 = v2t + 1/2 at^2
v1 = -v1* 2.40 + 1/2 * -9.8 * 2.40^2
v1 = -2.4v1 + -28.224
3.4v1 = -28.224
v1 = -8.301

check:
v1 = v2t + 1/2 at^2
-8.301 = 8.301 * 2.4 + 1/2 * -9.8 * 2.4^2
-8.301 = 19.9224 + -28.224
-8.301 = -8.301

Now, knowing that you threw the ball at 8.301 m/s the first time, construct a right triangle, where the opposite side is the vertical component of the velocity, the angle is 44.0 degrees, and the hypotenuse is the velocity that you throw the ball the second time.

The vertical component of velocity on the second throw will equal the speed thrown on the first throw - causing it to reach the same height.

sin 44.0 = opposite/hypotenuse
sin 44.0 = 8.301/hypotenuse
0.695 = 8.301/hypotenuse
hypotenuse = 8.301 / 0.695
hypotenuse = 11.944 m/s (solution!)