Physics Test Projectile Motion
Name______
Physics test – projectile motion
1. An Alaskan rescue plane drops a package of emergency rations to a stranded hiker. The plane is traveling horizontally at 40.0m/s at a height of 1.00 x 102 m above the ground. (a) Where does the package hit the ground relative to the point at which it was released? (b) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?
2. A long jumper leaves the ground at an angle of 20.0 o to the horizontal and at a speed of 11.0 m/s. (a) How long does it take him to reach maximum height? (b) What is the maximum height? (c) How far does he jump? Disregard the motion of his arms and legs.
3. One of the fastest recorded pitches in major-league baseball, thrown by Billy Wagner in 2003, was clocked at 101.0 mi/h. If a pitch were thrown horizontally with this velocity, how far would the ball vertically by the time it reached home plate, 60.5 ft. away?
4. The best leaper in the animal kingdom is the puma, which can jump to a height of 4 m when leaving the ground at an angle of 45o. With what speed must the animal reach the ground to reach that height?
5. Tom the cat is chasing Jerry the mouse across the surface of a table 1.5 m above the floor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 5.0m/s. Where will Tom strike the floor?
6. A place kicker must kick a football from a point 36.0m (about 39 yards) from the goal, and the ball must clear the crossbar, which is 3.05m high. When kicked, the ball leaves the ground with a velocity of 20.0 m/s at an angle of 53o to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?
Object launched horizontally
x= vxt
y= vyt + 1/2 gt2
t2=2y/g vy is 0
Object Not launched horizontally
*Must find the x and y components of initial velocity.
vx=vi cos q
vy=vi sin q
y= vyt + 1/2 gt2
t= -2vy/g y is 0
R = vxt
Ymax= vyt + 1/2 gt2 where t = 1/2 total time
y=H= vy2 –vi2 / 2g