# Finding Time of Flight/Maximum Height/Horizontal Range

## Homework Statement

Find Time of Flight/Maximum height & horizontal range given Launch Angle and Initial Velocity.

Using Launch angles: 9, 27, 45, 63 & 81.
Initial Velocity: 25 m/s

## The Attempt at a Solution

I used sine/cosine to find the Horizontal Range & Maximum height.

For launch angle 9°:
Maximum Height: 25sin9 =
Horizontal Range: 25cos9 =

I don't know if It's correct, also, I do not know how to find the Time of flight.

Thanks.

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Simon Bridge
Homework Helper
That would not be correct.
if the launch angle were 9deg, then 25sin(9) would be the y-component of the velocity.

You need to divide the initial velocity into x and y components.
Since the launch angle varies, you just leave it as ##\theta##.

From there - use kinematics to write out the maximum height and range equations in terms of the acceleration (you know this) in each direction and the velocity components.
You'll have to include an extra time of flight (T) term which you eliminate to get the final relations.

It is an important discipline in physics to be able to write down equations without knowing what parts of them are.