What is the initial velocity of a ball thrown straight up?

AI Thread Summary
The discussion focuses on calculating the initial velocity of a ball thrown straight up, given that it reaches a maximum height in 5.78 seconds and experiences an acceleration due to gravity of 9.8 m/s². Multiple participants confirm that the initial velocity can be calculated using the formula V_initial = g * t, leading to a consistent result of 56.644 m/s. At the highest point of its trajectory, the ball's velocity is zero, reinforcing the calculations. The consensus among contributors is that the initial velocity is indeed 56.644 m/s. This calculation is crucial for understanding projectile motion in physics.
chemguy990
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A ball of unknown mass is thrown straight up.

*maximum height reached in 5.78 seconds
*acceleration due to gravity 9.8 m/s2

Calculate initial velocity in m/s

Would the initial velocity be 9.8 m/s2 x 5.78 s: 56.644 m/s?

Thats all the information given.
 
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chemguy990 said:
A ball of unknown mass is thrown straight up.

*maximum height reached in 5.78 seconds
*acceleration due to gravity 9.8 m/s2

Calculate initial velocity in m/s

Would the initial velocity be 9.8 m/s2 x 5.78 s: 56.644 m/s?

Thats all the information given.
that's what i got

v_y=v_{0y}+gt

-v_{0y}=gt
 
Not to be redundant with the other post but i got the same=P

V initial x (Acceleration)(time) = V final

At the highest point, the velocity is 0 so

V initial + (-9.8m/s^2)(5.78s) = 0

V initial = 56.644 m/s
 
Inonin said:
Not to be redundant with the other post but i got the same=P

V initial x (Acceleration)(time) = V final

At the highest point, the velocity is 0 so

V initial + (-9.8m/s^2)(5.78s) = 0

V initial = 56.644 m/s
no prob :) so that's 3 for 56.644 m/s!
 
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