A ball thrown vertically upward with a a speed of 19.6 m/s from the top of the tower and returns to the earth in 6 second. find the height of the tower.
Answer:
Given :-
- A ball thrown vertically upward with a speed of 19.6 m/s from the top of the tower and returns to the earth in 6 seconds.
To Find :-
- What is the height of the tower.
Formula Used :-
Second Equation Of Motion Formula :
where,
- h = Height
- u = Initial Velocity
- t = Time Taken
- g = Acceleration due to gravity
Solution :-
Given :
- Initial Velocity (u) = 19.6 m/s
- Time Taken (t) = 6 seconds
- Acceleration due to gravity (g) = - 9.8 m/s²
According to the question by using the formula we get,
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Verified answer
Answer:
58.8 metres is the required height of the tower .
Explanation:
According to the Question
It is given that ball thrown vertically upward with a a speed of 19.6 m/s from the top of the tower and returns to the earth in 6 second. we need to calculate the height of the tower . obviously the height of tower is equal to the day distance covered by the ball in given time interval.
So , using Kinematics Equation
Acceleration due to gravty is 9.8m/s²
where,
h denote height of tower
u denote initial velocity
t denote time taken
g denote acceleration due to gravity
substitute the value we get
➻ h = 19.6×6 + ½×(-9.8) × 6² {as accⁿ due to gravity is acting in. downward direction)
➻ h = 117.6+ ½ × (-9.8)×36
➻ h = 117.6+ (-9.8)×18
➻ h = 117.6 - (9.8×18)
➻ h = 117.8 - 176.4
➻ h = -58.8m
➻h = 58.8 (as height cannot negative and its show that the ball covered 58.8 m distance from the top of the tower).