Question

43. 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 (a) as seen from the truck, (b) as seen
from the road..​

Answer 1

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Given :

• Speed = 14.7 m/s

• Distance = 58.8 m

To find:-

• The speed

• Angle of projection

[Time = Distance/Speed]

Substitute the known values in above formula

⇒ 58.8/14.7

⇒ 4 sec

We have ..

• Final velocity (v) = 0 m/s

• Time (t) = 4/2 = 2 sec

• Acceleration due to gravity (g) = -10 m/s.

Substitute the known values in above formula

⇒ 0 = u + (-10)(2)

⇒ u = 20 m/s

∴ Speed of truck is 20 m/s.

b) As seen from the road, will be resultant of speed of truck and ball.

Now,

• Speed of truck = 14.7 m/s

• Speed of ball = 20 m/s

⇒ √(u² + v²)

⇒ √[(19.6)² + (14.7)²]

⇒ √(384.16) + (216.09)

⇒ √600.25

⇒ 24.5 m/s

Angle seen from road,

⇒ tan-¹ (19.6)/(14.7)

⇒ tan-¹ (4/3)

⇒ 53°

•°• The speed of ball is 24.5 m/s with an horizontal angleof 53° as seen through the road.

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