Find the velocity with which a ball is thrown up if it reaches the maximum height in 5sec. Also find the maximum height.?

Question

Answer

U=0 t=5s a=10m/s^2... Minimum height=s=ut+1/2at^2... =(1/2)×10×25m=125m... Velocity=v... v^2=u^2+2as.. v^2=2×10×125.. v=Root2500 m/s=50m/s....

Answer

Answer

Understanding the Problem:
To find the velocity with which a ball is thrown up and the maximum height it reaches in 5 seconds, we need to utilize the equations of motion for projectile motion.

Finding the Initial Velocity:
- Let's assume the initial velocity of the ball when thrown up is 'u' m/s.
- At the maximum height, the vertical component of the velocity becomes 0 m/s.
- Using the equation v = u + at, where v = final velocity, u = initial velocity, a = acceleration (in this case, acceleration due to gravity), and t = time.
- Here, final velocity v = 0 m/s, acceleration a = -9.81 m/s² (as the ball is moving against gravity), and time t = 5 seconds.
- Substituting the values, we get 0 = u - 9.81 * 5
- Solving the equation, we find the initial velocity 'u' of the ball.

Finding the Maximum Height:
- The maximum height 'h' can be determined using the equation h = ut + (1/2)at², where h = height, u = initial velocity, t = time, and a = acceleration.
- Substituting the values, we get h = ut + (1/2)at²
- Substituting the values of 'u' and 't', we can calculate the maximum height the ball reaches.

Conclusion:
By following these steps, we can determine the initial velocity of the ball when thrown up and the maximum height it reaches in 5 seconds. This process helps in understanding the principles of projectile motion and applying the equations of motion to solve such problems effectively.

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