How high does the ball rise?
How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)
How long does the ball remain in the air? (Hint: The time it takes the ball to rise equals the time it takes to fall.)
-
Vf = Vi + a * t
Where
Vi = initial velocity = 18 m/s
Vf = final velocity = - Vi = - 18 m/s
a = acceleration = -9.8 m/s²
t = time = ?
so
- 18 = 18 + (-9.8) * t
t = (-36) / (-9.8)
t = 3.67 s
The ball remains in the air for 3.67 s
It will reach its highest point in just half that time.
T = t / 2
T = 3.67 / 2
T = 1.84 s
The highest point it reaches will be
d = ½ ( Vi + Vf) * t
where
d = distance = ?
Vi = initial velocity = 18 m/s
Vf = final velocity = 0 m/s
t = time = T = 1.84 s
so
d = ½ ( 18 + 0 ) * 1.84
d = 16.56 m
Its highest point will be 16.56 m
Where
Vi = initial velocity = 18 m/s
Vf = final velocity = - Vi = - 18 m/s
a = acceleration = -9.8 m/s²
t = time = ?
so
- 18 = 18 + (-9.8) * t
t = (-36) / (-9.8)
t = 3.67 s
The ball remains in the air for 3.67 s
It will reach its highest point in just half that time.
T = t / 2
T = 3.67 / 2
T = 1.84 s
The highest point it reaches will be
d = ½ ( Vi + Vf) * t
where
d = distance = ?
Vi = initial velocity = 18 m/s
Vf = final velocity = 0 m/s
t = time = T = 1.84 s
so
d = ½ ( 18 + 0 ) * 1.84
d = 16.56 m
Its highest point will be 16.56 m
-
This question is to do with one dimensional projectile motion, so straight away it should spring to mind that you can use the constant acceleration equations.
For this trajectory, take the positive direction to by along the positive y-axis (upwards).
1.
If an object is thrown vertically upwards, then at the highest point in its motion, it will be instantaneously at rest (so its speed will be zero) and applying this allows you to find the answer to the first question.
The information given is:
- initial velocity = u = 18m/s
- final velocity = v = 0 m/s
- acceleration due to gravity = a = -9.81m/s²
The acceleration is negative because it is acting vertically downwards at all times.
- displacement of the ball = s = ? m
The projectile motion equation that contains all the information you need is:
v² = u² + 2as
re-arranging for s...
s = (v² - u²) / 2a
= (0² - 18²) / (2*-9.81)
≈ 16.5 m.
So the highest point the ball rises to is 16.5m.
2.
For the second question, you know from the hint given that the time it takes the ball to rise and then fall is equal to half the time taken for the ball to rise. So if you find the time it takes the ball to rise to its highest point, then two times this time will give you the time the ball has spent in the air.
For this trajectory, take the positive direction to by along the positive y-axis (upwards).
1.
If an object is thrown vertically upwards, then at the highest point in its motion, it will be instantaneously at rest (so its speed will be zero) and applying this allows you to find the answer to the first question.
The information given is:
- initial velocity = u = 18m/s
- final velocity = v = 0 m/s
- acceleration due to gravity = a = -9.81m/s²
The acceleration is negative because it is acting vertically downwards at all times.
- displacement of the ball = s = ? m
The projectile motion equation that contains all the information you need is:
v² = u² + 2as
re-arranging for s...
s = (v² - u²) / 2a
= (0² - 18²) / (2*-9.81)
≈ 16.5 m.
So the highest point the ball rises to is 16.5m.
2.
For the second question, you know from the hint given that the time it takes the ball to rise and then fall is equal to half the time taken for the ball to rise. So if you find the time it takes the ball to rise to its highest point, then two times this time will give you the time the ball has spent in the air.
12
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