Question
A ball is projected with an initial upward velocity component of 25 m/s and a horizontal velocity component of 30 m/s.
a) What are the position and velocity of the ball after 2 s and 4 s?
b) How long a time is required to reach the highest point?
c) How high is this point?
d) What is the time of flight?
e) What is the range?
a) What are the position and velocity of the ball after 2 s and 4 s?
b) How long a time is required to reach the highest point?
c) How high is this point?
d) What is the time of flight?
e) What is the range?
Answers
GPT 3.5
a)
After 2 seconds:
Vertical velocity = initial vertical velocity - (acceleration due to gravity x time)
Vertical velocity = 25 m/s - (9.81 m/s^2 x 2 s) = 25 m/s - 19.62 m/s = 5.38 m/s
Vertical position = initial vertical position + (initial vertical velocity x time) - (0.5 x acceleration due to gravity x time^2)
Vertical position = 0 + (25 m/s x 2 s) - (0.5 x 9.81 m/s^2 x 2^2) = 50 m - 19.62 m = 30.38 m
Horizontal position = initial horizontal position + (horizontal velocity x time)
Horizontal position = 0 + (30 m/s x 2 s) = 60 m
After 4 seconds:
Vertical velocity = 25 m/s - (9.81 m/s^2 x 4 s) = 25 m/s - 39.24 m/s = -14.24 m/s
(Note: negative sign indicates downward direction)
Vertical position = 0 + (25 m/s x 4 s) - (0.5 x 9.81 m/s^2 x 4^2) = 100 m - 78.48 m = 21.52 m
Horizontal position = 0 + (30 m/s x 4 s) = 120 m
b) To find the time to reach the highest point, we can use the formula for vertical velocity:
Vertical velocity = initial vertical velocity - (acceleration due to gravity x time)
0 = 25 m/s - 9.81 m/s^2 x time
time = 25 m/s / 9.81 m/s^2
time = 2.55 s
c) The height of the highest point can be found using the formula for vertical position:
Vertical position = initial vertical position + (initial vertical velocity x time) - (0.5 x acceleration due to gravity x time^2)
Vertical position = 0 + (25 m/s x 2.55 s) - (0.5 x 9.81 m/s^2 x 2.55 s)^2
Vertical position = 63.75 m
d) The time of flight can be found by doubling the time it took to reach the highest point:
Time of flight = 2.55 s x 2 = 5.1 s
e) The range can be found by multiplying the horizontal velocity by the time of flight:
Range = 30 m/s x 5.1 s = 153 m
Therefore,
a) After 2 seconds, the ball's position is at 60 m horizontally and 30.38 m vertically. After 4 seconds, the ball's position is at 120 m horizontally and 21.52 m vertically.
b) It takes 2.55 seconds to reach the highest point.
c) The highest point is at a height of 63.75 meters.
d) The time of flight is 5.1 seconds.
e) The range of the ball is 153 meters.
After 2 seconds:
Vertical velocity = initial vertical velocity - (acceleration due to gravity x time)
Vertical velocity = 25 m/s - (9.81 m/s^2 x 2 s) = 25 m/s - 19.62 m/s = 5.38 m/s
Vertical position = initial vertical position + (initial vertical velocity x time) - (0.5 x acceleration due to gravity x time^2)
Vertical position = 0 + (25 m/s x 2 s) - (0.5 x 9.81 m/s^2 x 2^2) = 50 m - 19.62 m = 30.38 m
Horizontal position = initial horizontal position + (horizontal velocity x time)
Horizontal position = 0 + (30 m/s x 2 s) = 60 m
After 4 seconds:
Vertical velocity = 25 m/s - (9.81 m/s^2 x 4 s) = 25 m/s - 39.24 m/s = -14.24 m/s
(Note: negative sign indicates downward direction)
Vertical position = 0 + (25 m/s x 4 s) - (0.5 x 9.81 m/s^2 x 4^2) = 100 m - 78.48 m = 21.52 m
Horizontal position = 0 + (30 m/s x 4 s) = 120 m
b) To find the time to reach the highest point, we can use the formula for vertical velocity:
Vertical velocity = initial vertical velocity - (acceleration due to gravity x time)
0 = 25 m/s - 9.81 m/s^2 x time
time = 25 m/s / 9.81 m/s^2
time = 2.55 s
c) The height of the highest point can be found using the formula for vertical position:
Vertical position = initial vertical position + (initial vertical velocity x time) - (0.5 x acceleration due to gravity x time^2)
Vertical position = 0 + (25 m/s x 2.55 s) - (0.5 x 9.81 m/s^2 x 2.55 s)^2
Vertical position = 63.75 m
d) The time of flight can be found by doubling the time it took to reach the highest point:
Time of flight = 2.55 s x 2 = 5.1 s
e) The range can be found by multiplying the horizontal velocity by the time of flight:
Range = 30 m/s x 5.1 s = 153 m
Therefore,
a) After 2 seconds, the ball's position is at 60 m horizontally and 30.38 m vertically. After 4 seconds, the ball's position is at 120 m horizontally and 21.52 m vertically.
b) It takes 2.55 seconds to reach the highest point.
c) The highest point is at a height of 63.75 meters.
d) The time of flight is 5.1 seconds.
e) The range of the ball is 153 meters.