A ball thrown vertically upwards with an initial speed of 24.5m/s.Calculate the greatest height it will stain.
A ball thrown vertically upwards with an initial speed of 24.5m/s.
1.calculate the greatest heith it will atain.
2.how long does it takes to reach it's maximum heigh?
3.What is its speed when it returns to the ground?
To calculate the greatest height the ball will attain, we need to find the maximum displacement or height reached by the ball.
The key equation to use in this case is the kinematic equation for vertical motion:
v² = u² + 2as
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (24.5 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s² for objects moving upward)
s = displacement or height reached
Plugging in the values into the equation, we have:
0² = (24.5 m/s)² + 2(-9.8 m/s²)s
Simplifying,
0 = 24.5² - 19.6s
Solving for s,
19.6s = 24.5²
s = (24.5²) / 19.6
s = 30.81 meters
Therefore, the greatest height the ball will attain is approximately 30.81 meters.
To calculate the greatest height the ball will attain, we first need to determine the time it takes for the ball to reach its maximum height. We can use the following kinematic equation to solve for the time:
Vf = Vi + at
In this equation:
- Vf is the final velocity (at the maximum height) which is zero as the ball momentarily stops moving at its highest point.
- Vi is the initial velocity of the ball, which is 24.5 m/s.
- a is the acceleration, which in this case is acceleration due to gravity, approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
- t is the time we need to find.
Rearranging the equation to solve for time, we get:
t = (Vf - Vi) / a
Substituting the values we have:
t = (0 - 24.5) / -9.8
t = 2.5 seconds
The ball takes 2.5 seconds to reach its maximum height. To find the maximum height, we can use another kinematic equation:
Δy = Vi*t + 0.5*a*t^2
In this equation:
- Δy is the change in height, which is the maximum height we want to find.
- Vi is the initial velocity, which is 24.5 m/s.
- a is the acceleration due to gravity, which is -9.8 m/s^2.
- t is the time taken, which is 2.5 seconds.
Substituting the values, we get:
Δy = 24.5*2.5 + 0.5*(-9.8)*(2.5)^2
Δy = 61.25 - 30.625
Δy = 30.625 m
Therefore, the greatest height the ball will attain is approximately 30.6 meters.