## Projectile Motion:- Variety of 2D Questions: Level Ground

Q.1. Ayoob threw a ball upwards at an angle of 33 degrees to the horizontal and west on level ground at a speed of 42 m/s.

a. What was the initial horizontal component of the ball's velocity when it was thrown?
b. What was the initial vertical component of the ball's velocity when it was thrown?
c. How long did the ball take to reach maximum height?
d. What was the maximum height reached?
e. What was the ball's velocity at maximum height?
f. What was the total time of flight?
g. What was the horizontal component of the ball's velocity as it hit the ground?
h. What was the vertical component of the ball's velocity as it hit the ground?
i. What was the speed of the ball as it hit the ground?
j. At what angle does the ball hit the ground?
k. What is the range (horizontal distance travelled) of the ball's flight? (nearest m).

Q.2. Ken kicked a football upwards at an angle of 13 degrees to the horizontal and south on level ground at a speed of 52 m/s.

a. What was the initial horizontal component of the ball's velocity when it was kicked?
b. What was the initial vertical component of the ball's velocity when it was kicked?
c. How long did the ball take to reach maximum height?
d. What was the maximum height reached?
e. What was the ball's velocity at maximum height?
f. What was the total time of flight?
g. What was the horizontal component of the ball's velocity as it hit the ground?
h. What was the vertical component of the ball's velocity as it hit the ground?
i. What was the speed of the ball as it hit the ground?
j. At what angle does the ball hit the ground?
k. What is the range (horizontal distance travelled) of the football's flight? (nearest m).

Q.1. Ayoob threw a ball upwards at an angle of 33 degrees to the horizontal and west on level ground at a speed of 42 m/s.

a. What was the initial horizontal component of the ball's velocity when it was thrown?
Horizontal component = v cosθ = 42 cos 33 = 35.22m/s

b. What was the initial vertical component of the ball's velocity when it was thrown?
Vertical component = v sinθ = 42 sin 33 = 22.87m/s

c. How long did the ball take to reach maximum height?
v = u + at
0 = -22.87 = 9.8t t = 22.87/9.8 = 2.33s

d. What was the maximum height reached?
v2 = u2 + 2 a s
02 = (-22.87)2 + 2 x 9.8 x s
0 = 523.04 + 19.6s
s = 523.04/19.6
s = 26.69m

e. What was the ball's velocity at maximum height?
35.22m/s horizontally. Horizontal component of velocity is constant, vertical component = zero.

f. What was the total time of flight?
Time to reach maximum height = 2.33s
Total time of flight = 2 x 2.33 = 4.66s (time to go up = time to fall down)

g. What was the horizontal component of the ball's velocity as it hit the ground?
35.22m/s horizontally. Horizontal component of velocity is constant.

h. What was the vertical component of the ball's velocity as it hit the ground?
v2 = u2 + 2 a s
s = max height = 26.69m
v2 = 02 + 2 x 9.8 x 26.69
v2 = 523.124
v = √523.124 = 22.87m/s

i. What was the speed of the ball as it hit the ground?
speed = magnitude of final velocity = vector sum of vertical and horizontal velocities
Since vertical and horizontal velocities are at right angles, then Pythagoras' theorem may be applied.
Vertical component of final velocity = 22.87m/s
Horizontal component of final velocity = 35.22
v2 = vx2 + vy2
v2 = 35.222 + 22.872
v2 = 1240.45 + 523.04 = 1763.5
v = 41.99m/s

j. At what angle does the ball hit the ground?
tanθ = vertical velocity/horizontal velocity
= 22.87/35.22 = 0.649347 = 32.99deg = 33deg to horizontal.(nearest deg)

k. What is the range (horizontal distance travelled) of the ball's flight? (nearest m).

Range = horizontal velocity x time of flight
Range = 35.22 x 4.66 = 164.125m = 164m (nearest m)

Q.2. Ken kicked a football upwards at an angle of 13 degrees to the horizontal and south on level ground at a speed of 52 m/s.

a. What was the initial horizontal component of the ball's velocity when it was kicked?
Horizontal component = v cosθ = 52 cos 13 = 50.67m/s

b. What was the initial vertical component of the ball's velocity when it was kicked?
Vertical component = v sinθ = 52 sin 13 = 11.70m/s

c. How long did the ball take to reach maximum height?
v = u + at
0 = -11.70 = 9.8t
t = 11.70/9.8 = 1.19s

d. What was the maximum height reached?
v2 = u2 + 2 a s
02 = (-11.70)2 + 2 x 9.8 x s
0 = 136.89 + 19.6s
s = 136.89/19.6
s = 6.98m

e. What was the ball's velocity at maximum height?
50.67m/s horizontally. Horizontal component of velocity is constant, vertical component = zero.

f. What was the total time of flight?
Time to go up = time to come down
Total time of flight = 2 x 1.19 = 2.38s

g. What was the horizontal component of the ball's velocity as it hit the ground?
50.67m/s horizontally. Horizontal component of velocity is constant, vertical component = zero.

h. What was the vertical component of the ball's velocity as it hit the ground?
v2 = u2 + 2 a s
s = max height = 6.98m v2 = 02 + 2 x 9.8 x 6.98
v2 = 136.808
v = √136.808 = 11.70m/s

i. What was the speed of the ball as it hit the ground?
speed = magnitude of final velocity = vector sum of vertical and horizontal velocities
Since vertical and horizontal velocities are at right angles, then Pythagoras' theorem may be applied.
Vertical component of final velocity = 11.70m/s
Horizontal component of final velocity = 50.67
v2 = vx2 + vy2
v2 = 50.672 + 11.702
v2 = 1240.45 + 523.04 = 1763.5
v = 41.99m/s

j. At what angle does the ball hit the ground?
tanθ = vertical velocity/horizontal velocity
= 11.70/50.67 = 0.230906 = 13.002deg = 13deg to horizontal.(nearest deg)

k. What is the range (horizontal distance travelled) of the football's flight? (nearest m).

Range = horizontal velocity x time of flight
Range = 50.67 x 2.38 = 120.59m = 121m (nearest m)