A projectile's launch speed is five times its speed at maximum height.Find launch angle.How do I find this?

Question

Adrian Ayala · Accepted Answer

#cos^-1(1/5) ~~ 78^circ#

If the angle of projection is #alpha# and the projection speed is #u#, the projectile starts off with the horizontal and vertical velocity components #u cos alpha# and #u sin alpha#, respectively. Of these, the horizontal component of the velocity is constant throughout the motion, while the vertical component keeps on changing at a constant rate #-g#.

At the topmost point of the path, the projectile's vertical velocity component vanishes - and thus its velocity at that point is exactly horizontal, and its magnitude is #u cos alpha#.

Thus $u = 5u cos alpha implies cos alpha = 1/5#

Adrian Ayala · Answer

undefined The launch angle ($\theta$) can be calculated using the formula:

$$ \theta = \arctan\left(\frac{v^2}{gR}\right) $$

where:
$v$ = launch speed,
$g$ = acceleration due to gravity (approximately 9.8 m/s²),
$R$ = range of the projectile.

In this case, since the launch speed is five times the speed at maximum height, you can express $v$ in terms of the speed at maximum height ($v_{\text{max}}$):

$$ v = 5 \cdot v_{\text{max}} $$

Substitute this expression for $v$ into the launch angle formula to find the launch angle.