A ball is projected vertically up word from the top of tower 60m ,hight with velocity 30m ,what is the mex heigh above the ground leve? How long it take to reach the ground
For upward decelerated motion
v=Vo-gt
Since v=0, Vo=gt.
Let g= 10 m/s2,
t=Vo/g=30/10=3 s.
h=Vo•t-(gt^2)/2 =
=30•3-10•9/2=45 m.
Total height
H=45+80=125 m.
H=gt^2/2,
t=sqroot(2H/g)=sqroot (2•125/10)=5 s.
rubbish
To find the maximum height above the ground level and the time it takes for the ball to reach the ground, we can use the kinematic equations of motion.
1. Finding the maximum height:
The initial vertical velocity of the ball (u) is 30 m/s, and it goes against gravity until it reaches its maximum height, where its final vertical velocity (v) becomes zero. The acceleration due to gravity (g) is approximately 9.8 m/s², assuming no air resistance.
We can use the formula:
v^2 = u^2 + 2as
Here,
- v = 0 (final velocity)
- u = 30 m/s (initial velocity)
- g = -9.8 m/s² (acceleration due to gravity, negative because it opposes the motion)
- s = maximum height above the ground level (which is what we're trying to find)
Substituting the values into the equation and solving for s:
0^2 = (30 m/s)^2 + 2 * (-9.8 m/s²) * s
900 m²/s² = 19.6 m/s² * s
s = 900 m²/s² / 19.6 m/s²
s ≈ 45.92 m
Therefore, the maximum height above the ground level is approximately 45.92 meters.
2. Finding the time taken to reach the ground:
We can use the formula:
v = u + gt
Here,
- v = 0 (final velocity is zero because it hits the ground and stops)
- u = 30 m/s (initial velocity)
- g = -9.8 m/s² (acceleration due to gravity, negative because it opposes the motion)
- t = time taken to reach the ground (which is what we're trying to find)
Substituting the values into the equation and solving for t:
0 = 30 m/s + (-9.8 m/s²) * t
-30 m/s = -9.8 m/s² * t
t = -30 m/s / -9.8 m/s²
t ≈ 3.06 s
Therefore, it takes approximately 3.06 seconds for the ball to reach the ground.