A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 21.0 m/s from a height of 9.0 m.
(a) How high does the ball rise from its original position?
(b) How long does it take to reach its highest point?
(c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is the ball's velocity when it returns to the level from which it started?
a) hmax=vy^2/2g
so 21^2/(2*9.8)= 22.5 m
b) t=sqroot 2h/g
sqrt of 2*22.5 /9.8 = 4.59s
c) you can use kinematics here
h=9+22.5 = 31.5
vi=0
a=9.8
you can use
d=1/2 gt^2 to find time
31.5=(.5*9.8)t^2
t=2.535s
d) then use vf =vi+at
where vf=? vi=0 a=g=9.8 t=2.535
hope that helped
To solve these questions, we can use the equations of motion. In particular, we'll use the equations for motion under constant acceleration (in this case, the acceleration is due to gravity).
(a) To find the maximum height reached by the ball, we need to determine the time it takes to reach the highest point. We can use the equation:
v = u + at
where:
v is the final velocity (at the highest point, it will be 0 because the ball momentarily stops before falling back down)
u is the initial velocity (21.0 m/s)
a is the acceleration due to gravity (-9.8 m/s^2, negative because the acceleration is in the opposite direction to the initial velocity)
t is the time taken
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (0 - 21.0) / -9.8
t ≈ 2.142 seconds
Now that we have the time taken to reach the maximum height, we can use another equation:
s = ut + (1/2)at^2
where:
s is the displacement (maximum height) from the initial position (9.0 m)
u is the initial velocity (21.0 m/s)
a is the acceleration (-9.8 m/s^2)
t is the time taken (2.142 seconds)
Plugging in the values, we get:
s = (21.0 * 2.142) + (0.5 * -9.8 * 2.142^2)
s ≈ 22.45 meters
Therefore, the ball rises to a height of approximately 22.45 meters from its original position.
(b) The time taken to reach the highest point is the same as the time we calculated in part (a), which is approximately 2.142 seconds.
(c) To find the time it takes for the ball to hit the ground after reaching its highest point, we need to double the time taken to reach the highest point. This is because the time taken to go up and come down is the same. Therefore, the time taken to hit the ground is approximately 2 * 2.142 = 4.284 seconds.
(d) When the ball returns to the level from which it started, its velocity will be the same as its initial velocity, but in the opposite direction. So the velocity will be -21.0 m/s. The negative sign indicates that it's moving downward.