Question

The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 m/s, what is the direction of the initial velocity? (g=9.8 m/s2 )

Answer 1

as h= u^2 sin^2© /2g
where © is the angle of projection or direction of initial velocity with the horizontal
so, 10=28×28×sin^2© /2×9.8
so, sin^2©= 0.25
so, sin ©= 0.5
so, © =30 degree from horizontal....
thanks

Answer 2

An item or particle that is projected near the Earth's surface and travels along a curved route under the influence of gravity solely is said to be in projectile motion. Galileo demonstrated that this curved route is a parabola, although it might also be a line in the particular situation when it is hurled directly upwards.

For projectile motion, the maximum height is given by the formula,

h = $\frac{u^{2} sin^{2}\alpha }{2g}$

where,

h = maximum vertical height reached by the projectile.

u = initial velocity

α = angle with the horizontal at which the projectile is launched.

g = acceleration due to gravity.

To find, α.

Substituting given values in the equation,

10 = $\frac{28^{2} sin^{2}\alpha }{2*9.8}$

∴  $sin^{2}\alpha$ = $\frac{196}{784}$

∴ sinα = $\sqrt{\frac{196}{784} }$

∴ sinα = $\frac{14}{28}$

∴ sinα = $\frac{1}{2}$

∴ α = 30° wrt to horizontal.