A ball is thrown vertically upward with the velocity of 25m/s from the...
Problem Statement:
A ball is thrown vertically upward with the velocity of 25m/s from the top of tower of height 30m how long it will travel before it hit the ground?
Solution:
Step 1: Analyzing the given information
Before we start solving the problem, let's analyze the given information:
- Initial velocity (u) = 25m/s
- Height of the tower (h) = 30m
- Acceleration due to gravity (g) = -9.8m/s^2 (negative as it is acting downwards)
Step 2: Finding the time taken to reach maximum height
When the ball is thrown vertically upward, it will reach a maximum height and then start falling downwards due to the force of gravity. At the maximum height, the velocity of the ball becomes zero. We can find the time taken to reach the maximum height using the following formula:
Final velocity (v) = u + gt
We know that at maximum height, final velocity (v) = 0. So, we can substitute v=0 in the above formula and find the time (t) taken to reach the maximum height:
0 = 25 - 9.8t
t = 25/9.8 = 2.55 seconds
Step 3: Finding the time taken to fall from maximum height to the ground
From the maximum height, the ball falls freely under the influence of gravity. We can find the time taken to fall from maximum height (h) to the ground using the following formula:
h = ut + 1/2 gt^2
Substituting the given values, we get:
30 = 0 + 1/2 (-9.8) t^2
t^2 = 30/-4.9
t^2 = 6.12
t = sqrt(6.12) = 2.47 seconds
Step 4: Finding the total time taken
The total time taken by the ball to travel from the top of the tower to the ground is the sum of the time taken to reach the maximum height and the time taken to fall from the maximum height to the ground:
Total time taken = time taken to reach maximum height + time taken to fall from maximum height to the ground
Total time taken = 2.55 + 2.47 = 5.02 seconds
Conclusion:
The ball will travel for 5.02 seconds before it hits the ground.