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

Answer:

Velocity of truck v

t

​

=14.7m/s

Distance s=58.8m

(a)Since the ball returns back to its initial; position,angle of projection as seen from the truck =90

∘

Time taken by truck :t=

vt

s

​

=

14.7

58.8

​

=4s

t=

g

2usinθ

​

u=

2sinθ

tg

​

=

2×1

4×9.8

​

=2×9.8=19.6m/s

(b)As een from the road: v= u x2 +u y2 = v t2 +u 2= (4.7) 2+(9.6) 2

​

=24.5m/s

Angle of θ=tan −1 ( 14.719.6 )=tan −1

( u xu y)=tan −1 (1.33)=53

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