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A Person is Standing on a Truck Moving with a Constant Velocity of 14.7 M/S on a Horizontal Road. the Man Throws a Ball in Such a Way that It Returns to the Truck After the Truck Has Moved 58.8 M.

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प्रश्न

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection as seen from the truck .

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उत्तर

Given:
Velocity of the truck = 14.7 m/s
Distance covered by the truck when the ball returns again to the truck = 58.8 m

Therefore, we can say that the time taken by the truck to cover 58.8 m distance is equal to the time of the flight of the truck.
Time in which the truck has moved the distance of 58.8 m:

\[\text{ Time }(T) = \frac{s}{v} = \frac{58 . 8}{14 . 7} = 4 s\]
We consider the motion of the ball going upwards.
T = 4 s
Time taken to reach the maximum height when the final velocity v = 0:
\[t = \frac{T}{2} = \frac{4}{2} = 2 s\] 
a = g = −9.8 m/s2 (Acceleration due to gravity)
∴ v = u − at
⇒ 0 = u + 9.8 × 2
⇒ u = 19.6 m/s           
19.6 m/s is the initial velocity with which the ball is thrown upwards. 

 

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पाठ 3: Rest and Motion: Kinematics - Exercise [पृष्ठ ५३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 3 Rest and Motion: Kinematics
Exercise | Q 43.1 | पृष्ठ ५३

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