Question

5.
A toy plane p starts flying from point A along a straight
horizontal line 20 m above ground level starting with zero initial
velocity and acceleration 2 m/s2 as shown. At the same instant,
a man P throws a ball vertically upwards with initial velocity
'u'. Ball touches (coming to rest) the base of the plane at point
B of plane's journey when it is vertically above the man. 's' is
the distance of point B from point A. Just after the contact of
ball with the plane, acceleration of plane increases to 4 m/s2.
Find:
(0)
(ii)
(iii)
Initial velocity 'u' of ball.
Distance 's'.
Distance between man and plane when the man
catches the ball back. (g = 10 m/s2)​

Answer 1

Explanation:

Time taken by ball to reach B from O=t= times taken by plane to each B from A

(i) speed of ball at B=O=V

v

2

=u

2

−2gs⇒U

2

=2×10×20⇒

u=20m/s

(ii) v=u−gt⇒u=20−(10t)⇒

t=2sec

for plane S=ut+

2

1

at

2

=0+

×2×2

2

=4m

⇒

s=4m

(iii) time taken by ball to return back =2sec

Let BC=d

speed of plane at B=u+at=0+(2×2)=4m/s

d=(4×2)→

2

1

×3×2

2

=8+6=14m

distance between man and plane =

20

2

+14

2

=24.4m

solution