A projectile is fired with an initial speed of 31.0 at an angle of 10.0 above the horizontal. The object hits the ground 7.50 later.

Question

a. How much higher or lower is the launch point relative to the point where the projectile hits the ground?

b. To what maximum height above the launch point does the projectile rise?

c. What is the magnitude of the projectile's velocity at the instant it hits the ground?

d. What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?

Answer 1

With projectiles fired at an angle, we consider the vertical component (vy=v0 sin(θ)) and horizontal component (vx=v0 cos(θ)) separately.

vx stays constant throughout.
vy can be treated like a free fall after an iniital velocity of vy.

(a)
So assuming the projectile is fired at y=0, then calculate time t to reach maximum height attained using
t=vy/g
Subtract from total time T of 7.5 seconds to get the time of descent t1.
The distance descended, S1 is given
by
S1 = 0*t1+(1/2)(-g)t1²
Take the difference of S and S1 to get the difference of elevations. Watch the signs.

(b)
use S=(0-vy²)/(2(-g))

(c)
The vertical (downwards) velocity at impact is given by
vy1=t1*(-g)
Add this vectorially to vx to get the velocity.

(d)
the angle θ below the horizontal can be obtained by the arctangent of the ratio of vertical and horizontal components of velocity, namely:
tan(θ) = vy1/vx
(watch the signs, &theta should be between 0 and 90 degrees).