Calculating Time of Flight & Velocity of a Ball Thrown Upwards

[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

 

[Herman Trivilino]

You haven't considered the horizontal motion.

 

[crism7]

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

??

 

[PeroK]

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 $v = 8.02 m/s$?

 

[crism7]

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

vi = v1y / sin55
vi = -6.566 / 0.819152
vi = 8.02

 

[PeroK]

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.

 

[crism7]

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?

 

[PeroK]

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 $v_i = 6.6 \ m/s$.

 

[crism7]

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?

 

[PeroK]

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.

because the horizontal and vertical velocities are different?

They depend on the angle.

and then find the horizontal velocity with vi cos 55, and then solve for time of flight using t = d/v?

Yes.

 

[crism7]

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!

 

[PeroK]

t = 1.09

therefore the time of flight is 1.2s?

This confuses me. $1.09 \ne 1.2$

 

[crism7]

1.1 :']

 

[berkeman]

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.

 

[berkeman]

And a descriptive thread title has been restored on your thread to help any searches by your instructors. Have a nice day.