# Calculating Time of Flight & Velocity of a Ball Thrown Upwards

• crism7
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crism7
Homework Statement
.
Relevant Equations
.
Vertical components:
dy = 0m
ay = 9.8m/s^2 [down]
t = 1.34s
V1y = required
V2y = 0

i first tried to find V1y
dy =vi t + 1/2 a t^2
and got V1y = -6.566

then i solved for time of flight
dy =vi t + 1/2 a t^2
0 = -6.566t + 4/9t^2
and for 1.34 seconds

does this mean the time of flight is the same when the ball is thrown directly upwards??
is this correct, I'm not sure where to begin when solving this problem

You haven't considered the horizontal motion.

bob012345
Mister T said:
You haven't considered the horizontal motion.
dx = 4.14 m
Vx = 8.02cos55 = 4.6 m/s
t = required

t = d/v
t = 4.14 / 4.6
t = 0.9 s

??

crism7 said:
dx = 4.14 m
Vx = 8.02cos55 = 4.6 m/s
t = required

t = d/v
t = 4.14 / 4.6
t = 0.9 s

??
How did you get ##v = 8.02 m/s##?

PeroK said:
How did you get ##v = 8.02 m/s##?
vi = v1y / sin55
vi = -6.566 / 0.819152
vi = 8.02

crism7 said:
vi = v1y / sin55
vi = -6.566 / 0.819152
vi = 8.02
Okay, if you fire the projectile straight up, then the initial velocity is ##6.6 \ m/s##. That's correct.

And, you are assuming that for any angle, the vertical component of velocity is ##6.6 \ m/s##?

The problem with that is if you fire the projectile at ##1## degree above the horizontal and its vertical component remains ##6.6 \ m/s##, then its total speed is ##435 m/s##, which is faster than the speed of sound! Unlikely for a foam popper!

You should assume instead that the total velocity is the same for the popper and the horizontal and vertical components vary according to the angle of launch.

so the ball thrown upwards and the foam ball launched at an angle must have the same initial velocity of 6.6m/s?

dx = 4.14 m
Vx = 6.6m/s
t = required

and then solve for t?

crism7 said:
so the ball thrown upwards and the foam ball launched at an angle must have the same initial velocity of 6.6m/s?

dx = 4.14 m
Vx = 6.6m/s
t = required

and then solve for t?
The first time you had ##v_y = 6.6 \ m/s##. This time you have ##v_x = 6.6 \ m/s##. Neither of those represent ##v_i = 6.6 \ m/s##.

PeroK said:
The first time you had ##v_y = 6.6 \ m/s##. This time you have ##v_x = 6.6 \ m/s##. Neither of those represent ##v_i = 6.6 \ m/s##.
i need to find Vi? because the horizontal and vertical velocities are different? and then find the horizontal velocity with vi cos 55, and then solve for time of flight using t = d/v?

crism7 said:
i need to find Vi?
You already did. And, I've confirmed it. ##v_i = 6.6 \ m/s##. That's how fast the foam popper fires things.
crism7 said:
because the horizontal and vertical velocities are different?
They depend on the angle.
crism7 said:
and then find the horizontal velocity with vi cos 55, and then solve for time of flight using t = d/v?
Yes.

t = 4.14 m / 6.6cos55
t = 4.14 / 3.785604
t = 1.09

therefore the time of flight is 1.2s?
also, thank you so much for your help!

crism7 said:
t = 1.09

therefore the time of flight is 1.2s?
This confuses me. ##1.09 \ne 1.2##

1.1 :']

PeroK
crism7 said:
Homework Statement:: .
Relevant Equations:: .

Vertical components:
dy = 0m
ay = 9.8m/s^2 [down]
t = 1.34s
V1y = required
V2y = 0

i first tried to find V1y
dy =vi t + 1/2 a t^2
and got V1y = -6.566

then i solved for time of flight
dy =vi t + 1/2 a t^2
0 = -6.566t + 4/9t^2
and for 1.34 seconds

does this mean the time of flight is the same when the ball is thrown directly upwards??
is this correct, I'm not sure where to begin when solving this problem
It appear that you have a habit of deleting your thread title after you have received help on your homework. This thread is now locked. Please check your PMs -- we do not tolerate that type of behavior in our Homework Help forums.

## 1. How do you calculate the time of flight for a ball thrown upwards?

The time of flight for a ball thrown upwards can be calculated using the formula:
t = 2 * v0 / g
Where t is the time of flight, v0 is the initial velocity, and g is the acceleration due to gravity (9.8 m/s2).

## 2. What is the initial velocity of a ball thrown upwards?

The initial velocity of a ball thrown upwards is the velocity at which the ball is thrown. This can be measured using a speedometer or by calculating the change in position over time.

## 3. How do you calculate the velocity of a ball thrown upwards at a specific time?

The velocity of a ball thrown upwards at a specific time can be calculated using the formula:
v = v0 - g * t
Where v is the velocity at the specific time, v0 is the initial velocity, and g is the acceleration due to gravity (9.8 m/s2).

## 4. Can the time of flight and velocity of a ball thrown upwards be affected by air resistance?

Yes, air resistance can affect the time of flight and velocity of a ball thrown upwards. This is because air resistance creates a force that opposes the motion of the ball, causing it to slow down and potentially altering its trajectory.

## 5. How does the mass of the ball affect the time of flight and velocity when thrown upwards?

The mass of the ball does not affect the time of flight and velocity when thrown upwards. This is because the force of gravity and the acceleration due to gravity are constant, regardless of the mass of the object. However, a heavier ball may experience more air resistance, which can slightly alter its time of flight and velocity.

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