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 velocitu '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 4m/s². Find:

Question

(1) initial velocity 'u' of ball.
(2) distance 's'
(3) distance between man and plane when the man catches the ball back.(g=10 m/s²)

Answer 1

well, first do the vertical problem, the ball

how long does it take to fall 20 m?

20 = (1/2) (10) t^2
t^2 = 4
t = 2 seconds

so it will take to seconds to fall. By symmetry it will also take 2 seconds to rise and hit the plane

how fast is it going when it hits the ground?
u = a t
= 10*2 = 20 m/s
again by symmetry it will need 20 m/s initial velocity up to peak at 20 m

Now work on the horizontal problem, the plane.

it accelerates at 2 m/s^2 for 2 seconds
s = (1/2) a t^2 = (1/2)(2) (4)
s = 4 meters

its speed then is
v = a t = 2*2 = 4 m/s
it goes on for another 2 seconds, accelerating faster while the ball falls
d = Vi t + (1/2) a t^2
d = 4(2) + (1/2)(4)(4)
= 8 + 8
= 16
so
the horizontal distance between man and plane is 16 meters when he catches the ball
BUT
the plane is still 20 meters up
so
total distance = sqrt(16^2+20^2)
= 25.6 meters

Answer 2

Genius

Answer 3

I really like your logical explanation.
Thank you so much.

Answer 4

Could have been more precise

Answer 5

You need to work upon it.