Question

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 speed and the angle of projection a. as seen from the truck, b. as seen from the road

Answer 1

Answer:

Velocity=14.7 m/s 

Time taken by the truck to cover 58.8=t=S/v=58.8/14.7 =4sec

Since ball returns back to truck its angle of projection is vertically upwards with respect to truck.

u=speed of projection.

t=4s

h=0

h=4u-1/2 x 9.8x16

0=4u-78.4

4u=78.4

u=78.4/4=19.6 m/s

b) As seen from the road the speed of 

 ball will be resultant of 2 speeds vertical

speed =u=19.6 m/s

Horizontal speed given by truck=14.7m/s

⇒√u²+v²=√19.6²+14.7²=24.5 m/s

Thus angle seen from road:

Tan⁻¹[19.6/14.7] 

=Tan⁻¹1.33

=Tan⁻¹[4/3]

=53°with horizontal

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Answer 2

$\bold{\underline{\underline{Solution\::}}}$

Given :

• Speed = 14.7 m/s
• Distance = 58.8 m

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/s²

(Ball is going upward)

We know that..

v = u + at

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 angle of 53° as seen through the road.