Name: ______________________________________ Date: _____________ Period: _______
Mr. Talboo – Physics
Projectile Motion Practice Problems 2
1. A ball is thrown in such a way that its initial vertical and horizontal components of velocity
are 40 m/s and 20 m/s, respectively. Find the total time of flight and the distance the ball is from
its starting point when it lands (assume symmetrical trajectory)
t = (Vf – Vi) / -g = (-40 – 40) / -9.8 = 8.163 s
∆x = Vx t = (20 m/s) x 8.163 s = 163.265 m
2. A plane drops a package of supplies to a party of explorers, as shown in the figure below. If
the plane is traveling horizontally at 40.0 m/s and is 100 m above the ground, where does the
package strike the ground relative to the point at which it is released?
∆y = ½ gt2
t = √(-2 ∆y/g) = √(-2 (-100)/9.8) = 4.518s
∆x = Vx t = (40 m/s) (4.518s) = 180.702 m
3. A long-jumper leaves the ground at an angle of 20° above the horizontal and at a speed of
11.0 m/s.
(a) How long is he in the air?
(b) How far does he jump in the horizontal direction?
(c) What is his maximum height?
A) Vxi = 11 cos 20° = 10.337 m/s
Vyi = 11 sin 20° = 3.762 m/s
t = (Vf – Vi) / g = (-3.762 – 3.762) / -9.8 = 0.768 s
B) R = (Vi2 sin 2θ) / g = (112 sin 40) / 9.8 = 7.936 m
C) H = (Vi2 sin2θ) / 2g = (112 (sin 20)2) / 2(9.8) = 0.722 m
2. A fire hose ejects a stream of water at an angle of 29° above the
horizontal. The water leaves the nozzle with a speed of 35 m/s.
Assuming the water behaves like a projectile, how far from the
building should the fire hose be located to hit the highest possible fire?
Vxi = V cos θ = 35 cos 29 = 30.612 m/s
Vyi = V sin θ = 35 sin 29 = 16.968 m/s
t = Vf – Vi / g = 0 – 16.968 / -9.8 = 1.731 s
∆x = Vx t = (30.612)(1.731) = 52.989 m (greater amount of rounding error)
OR YOU COULD USE THE RANGE EQUATION, DIVIDING YOUR RANGE BY 2
SINCE YOU ARE LOOKING FOR MAX HEIGHT:
R = (Vi2 sin 2θ) / g = (352 sin 58) / 9.8 = 106.006 m / 2 = 53.003 m