A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?
1 answer:
Answer:
Speed, u = 29.4 m/s
Explanation:
Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s
Let u is speed with which the ball is thrown up. When the ball falls, v = 0
Using first equation of motion as :
v = u + at
Here, a = -g
So, u = g × t
u = 29.4 m/s
So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.
You might be interested in
Answer:
Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.
Explanation:
Both balls will hit the ground at the same distance and at the same time assuming they both start at the same initial elevation from the ground.
We know that range in the projectile motion given as
sin 60° = sin 120° =0.866
Therefore, the maximum range will be identical assuming both golfers hit the ball at the same elevation from the ground.
As for the time...
Where v is 0 m/s (final velocity at highest point), v0 is the initial velocity (for both golfers) and is the acceleration due to gravity (-9.8 m/s^²). As you can see, the time it takes to get to the highest point is independent of the angle; it is only dependent on the initial velocity. Since both golfers hit the ball with the same speed, the time for the ball to reach the highest point will be the same. Also, since it's a parabola, you can multiply the time by 2 for each golfer to get the time it takes to hit the ground.
Therefore, both balls will hit the ground at the same distance from their respective starting points and at the same time.