Hi, i'm kinda lost with this one.
a ball is thrown at 15m/s at an angle of 30 degrees to the horizontal , how far does it travel horizontally before making contact with the ground?
THANKS IN ADVANCE!
You'll need to determine how long the object is in the air first. Then the distance traveled is v*cos(theta)*t where v is the velocity. theta is the angle of elevation with the horizon and t is the time in the air. Right now you have the velocity and the angle, but you need to know the time.
To calculate the time use h=v*sin(theta)*t - (1/2)g*t^2
Set h to 0 and solve for t using the quadratic formula. You could also use the formula
v_f=v*sin(theta) - g*t and solve when v_f =0. That's the time needed to reach the max height; then double that time to get the time total. Again, v is the initial velocity and theta the angle of elevation.
Jackie: Roger is correct. Use the vertical component of velocity to find the time in air to the maximum height(at the top, velocity vertical is zero), then double that time to find total time in the air. Then, the horizontal distance is easy: distance=horizontalcomponent of initial velocity times the total time in the air .
Thanks!