Question

At what angle must a projectile be fired to travel 3.624m from a height of 1.012m?

x=3.624m
y=1.012m
v0=?
t=?
v=?

Answer 1

Here it is a approach of formula...
We know for the projectile motion the following formulae,
`H=(U^2sin^2theta)/2g` , `R=(U^2sin2theta)/g` , `t=(2Usintheta)/g`
Here `H` is the height of the projectile along `Y` axis and `R` is the range along `X` axis. `U` is the initial velocity and `theta` is the angle of projectile thrown.
Here we get , `H=1.012m` and `R=3.624m`

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