physics
posted by Jon on .
5)A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the hill would the ball hit the ground?
45.0 m
6)A missile launched at a velocity of 30.0 m/s at an angle of 30.0 to the normal. What is the maximum height the missile attains?
11.5 m

5)
I did not get the same answer you got.
Since there is no vertical component, the time to hit the ground is the same as if th ball had been dropped from a height of 50.0m. Calculate how long it takes for the ball to fall 50.0m. Take that time and multiply by the horizontal velocity (given as 10.0 m/s).
6)
I did not get the same answer you had.
Again, break the problem into vertical and horizontal components. Find the initial vertical component (30.0 m/s) * sin(60) because the problem says the missile is launched 30 degress from the NORMAL.
Determine how long before the vertical velocity is 0. That will be the maximum height. Plug that time into the distance equation:
t = time
g=gravity acceleration
distance = (initial velocity) * t + (1/2)(g)(t^2)
Remember that for positive up, gravity will be a negative acceleration.
So, pay attention to the sign of the second term.