A ball thrown vertically upward with a speed 40 m/s from top of a tower of height 45m
Answers
Answer:
3m
Explanation:
Firstly, ball is thrown upwards,
Initial Velocity = 10 m/s.
Final Velocity = 0
Acceleration (a) = -g
= -10 m/s² [Since, the Ball is thrown against the gravity.
∴ v² - u² = 2aS
(0) - (10)² = 2aS
2(-10)S = -100
S= 100/20
S = 5 m.
Height to which the ball reaches from the top of the tower is 5 m.
∴ Height to which the ball reaches from the surface of the earth = (40 + 5)m.
= 45 m.
Now, Time taken by the ball to reach the height of 5 m from the top of the tower is given by the formula,
∵ v - u = at
∴ 0 - 10 = (-10)(t)
∴ t = 1 seconds.
We know, the time of Ascent = Time of Descent.
∴ Time taken by the ball to again reach the top of the tower = 1 seconds.
When the ball starts falling from the height of 45 m.
Now, At the Height of 45 m,
Initial Velocity of the ball = 0
Final Velocity = v
Time taken to reach the top of tower = 1 seconds.
Acceleration due to gravity = 10 m/s²
Firstly, we will find the Final Velocity of the ball, after which we can calculate the total time.
∴ Height = 5 m [From the to of the tower is taken]
∴ v² - u² =2ah
v² - (0)² = 2(10)(5)
v² = 100
∴ v = 10 m/s.
Hence, the velocity by which the ball is falling is 10 m/s.
Now,
For Finding the total time,
Height of the ball (H) = 45 m.
Initial Velocity of the ball (u) = 0 m/s.
[∵ The ball is under the free fall].
Final Velocity of the ball (v) = 10 m/s.
Acceleration = 10 m/s²
Now, Using the Formula,
S = ut + 1/2 at²
45 = (0)(t) + 1/2 × 10 (t)²
t² = 45/5
t² = 9
t = 3 seconds.
Hence, the total time taken by the ball to strikes the ground from the 45 m height is 3 m.
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