v0

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1 2 3

Name: __________________________

Motion of Projectiles – Practice

1. A ball is launched into the air at the position shown on the left. Assume no friction or air drag. Draw where the ball will land when the cart reaches the end of the track.

A. Level Track/Constant Velocity

B. Level Track/Acceleration

C. Inclined Track/Acceleration

2. A football is kicked from the ground level. Ignore the effects of air on the flight of the ball. Rank the paths from greatest to least according to:

A. Time of flightB. Initial vertical velocity componentC. Initial horizontal velocity componentD. Initial speed

3. Beginning with the kinematic equations for projectiles,x displacement: x – x0 = vx0t = (v0cos 0)ty displacement: y – y0 = vy0t + ½gt2 = (v0sin 0)t + ½gt2

(where g = -9.80 m/s2), derive expressions for the:

A. Time to reach the maximum height for a symmetric projectile. [Hint: vy = 0 at ymax]

B. Maximum height of a symmetric projectile.

C. Total flight time for a symmetric projectile.

D. Range of a symmetric projectile.

4. A rifle that shoots bullets at 460 m/s is to be aimed horizontally at a target 45.7 m away. How high above the center of the target must the rifle barrel be held so that the bullet hits dead center?

5. A football kicker gives the ball an initial speed of 25 m/s at an angle of 28o. The goal post is 50 m away and its horizontal cross bar is 3.44 m above the ground. Does the kicker make the field goal? How much above or below the bar does the ball pass?

6. In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over three Ferris wheels and into a net. Assume that he is launched with a speed of 26.5 m/s and at an angle of 53.0° and neglect air drag.

A. Treating him as a particle, calculate his clearance over the first wheel.

B. If he reached maximum height over the middle wheel, by how much did he clear it?

C. How far from the cannon should the net's center have been positioned?

7. A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in the graph below where t = 0 is the instant the ball is struck. The scaling on the vertical axis is set by va = 19 m/s and vb = 31 m/s.

A. How far does the golf ball travel horizontally before returning to ground level?

B. What is the maximum height above ground level attained by the ball?

Solutions:1. A. On the cart B. Short of the cart C. On the cart2. A. All the same B. All the same C. 3, 2, 1 D. 3, 2, 1

3. A. t yMax =

v o sin θog

B. yMax = H =

(v o sin θo )2

2g

C. t tot = T =

2vo sin θog

D. xMax = R =

vo2 sin 2θog

4. 4.84 cm5. No. 1.99 m below6. A. 5.3 m B. 7.9 m C. 69 m7. A. 95 m B. 31 m