Question

A projectile is fired with an initial speed of 33 m/s at an angle of 57° above the horizontal. The object hits the ground 8.2 s later. (Neglect air resistance.)

(a) How much higher or lower is the launch point relative to the point where the projectile hits the ground? (If the point is lower, enter a negative number.)

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

(c) What are the magnitude and direction of the projectile's velocity at the instant it hits the ground?

Answer 1

in reply to a. a hint:
determine vertical speed
wonder if it is possible with this initial speed that the projectile returns in 8.2 s to its launching point.

Answer 2

vertical speed= 33 * sin(57)=27.7 m/s
it takes for the projectile t=v/a = 27.7 / 9.81 = 2.82 s to slow down to zero.
There is 8.2-2.82 = 5.18 s time left to fall down.
\[s = 1/2 \times a \times t ^{2} = 1/2 \times 9.81 \times 5.18^{2} = 131.6 m\] (answer b)

How much higher or lower is the platform? We need to know how much distance the projectile makes upwards:
\[s = 1/2 \times a \times t^{2} = 1/2 \times 9.81 \times 2.82^{2} = 39 m\]
The platform will be on 131.6-39 = 92.6 meters above the ground. (answer a)

c. determine vertical velocity and determine horizontal velocity ( 33 x cos(57) )
and combine these two.