a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point
Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.
(a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point )
v = Vi - g t
0 = Vi - 9.8 t where t is time to stop at top
so
Vi = 9.8 t
32 = 0 + Vi t - 4.9 t^2
32 = 9.8 t^2 - 4.9 t^2
t^2 = 6.53
t = 2.56 seconds to top
so
Vi = 9.8*2.56 = 25 m/s
8 = 0 + 25 t - 4.9 t^2 where ti is time at 8 m
4.9 t^2 -25 t + 8 = 0
t^2 - 5.11 t + 1.63 = 0
t = [ 5.11 +/- sqrt(26.11-6.52) ]/2
= [ 5.11 +/- 4.43]/2
small answer is on the way up, use it
t = .34 seconds
v = 25 - 9.8*.34 = 21.7 m/s
To find the velocity of the ball when it is 8m above the launch point, we can use the principles of motion and kinematics.
First, let's define the variables:
- Initial velocity (u) = unknown
- Final velocity (v) = ?
- Acceleration (a) = acceleration due to gravity (-9.8 m/s^2, considering downward as negative)
- Displacement (s) = 8m (the distance above the launch point)
- Maximum height (h) = 32m
We need to find the final velocity (v) at a certain height (s).
Since the ball was thrown upwards and reached a maximum height of 32m, we know that the final velocity at the maximum height is 0m/s, as it is momentarily at rest before it starts coming back down. This gives us a reference point.
From this information, we can use the equation of motion:
v^2 = u^2 + 2as
Plugging in the values, we have:
0^2 = u^2 + 2(-9.8)(32)
Simplifying the equation:
0 = u^2 - 627.2
Solving for u:
u^2 = 627.2
u ≈ √627.2
u ≈ 25 m/s (approx.)
Therefore, the initial velocity (u) of the ball is approximately 25 m/s, when it is 8m above the launch point.