A stone is thrown vertically upward from the top of a tower of height 24.5m with an initial velocity of 19.6 m/s. Calculate :-
➻ (a) the height to which the stone will rise before
returning to the ground
➻(b) the velocity with which it will strike the ground
➻(c) the total time of its journey.
Answers
According to the Question
It is given that stone is thrown vertically upward from a tower of height 24.5m.
- Initial velocity ,u = 19.6m/s
- Acceleration due to gravity ,g = 9.8m/s
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(a) The height to which the stone will rise before
returning to the ground .
In this case the stone attained its maximum position. So, the final velocity of the stone will be 0m/s.
Now, Calculating height attained by stone before returning to the ground. Using Kinematics Equation
- v² = u² + 2gh
substitute the value we get
➻ 0² = 19.6² + 2×(-9.8) ×h
➻ 0 = 384.16 + (-19.6)*h
➻ -384.16 = - 19.6*h
➻ -384.16/-19.6 = h
➻ h = 384.16/19.6
➻ h = 19.6m
- Hence, the height attained by stone before returning to the ground will be 19.6 metres.
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(b) the velocity with which it will strike the ground.
In this case we have to calculate the final velocity of the stone and the initial velocity will be zero .As the stone reached its maximum height . So, the total height covered by stone = height of tower + height attained by stone
= 24.5m + 19.6
= 44.1m
Using Kinematics Equation
- v² = u² + 2gh
Substitute the value we get
➻ v² = 0² + 2×9.8 × 44.1
➻ v² = 0 + 19.6×44.1
➻ v² = 864.36
➻ v = √864.36
➻ v = 29.4m/s
- Hence, the velocity which it will strike the ground will be 29.4m/s.
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(c) the total time of its journey.
In this case we have to calculate the total time taken in this journey . Using Kinematics Equation.
- h = ut + 1/2gt²
substitute the value we get
➻ 44.1 = 0×t + 1/2 × 9.8 × t²
➻ 44.1 = 0 + 4.9 × t²
➻ 44.1 = 4.9 × t²
➻ 44.1/4.9 = t²
➻ 9 = t²
➻ t = √9
➻ t = 3s
- Hence, the total time taken of this journey will be 3 second .
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