The starting hight is .51 metres above the ground, and the range is 4.651 metres. the angle of launch is 22 degrees above the horazontal. i need to find the initial velocity, and this was all i have to work with.
In a problem where the net acceleration is in a direction opposite to the positive direction you chose, the "a" in your formula becomes negative.I couldn't help but notice...
Shouldn't "-0.51 = vsin22t - 0.5gt^2" be -0.51 = vsin22t + 0.5gt^2?
The formula is d = Vit + 1/2 at^2 is it not?