Grade 12Mechanics

A man in a balloon rising vertically with an acceleration of 4.9m/sec2 , releases a ball 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is
A)14.7m
B)19.6m
C)9.8m
D)24.5m

Kaviya

8 Years agoGrade 12

2 Answers

Arun

Approved Tutor Answer8 Years ago

Take upward direction as positive.
Let v0 = velocity of the ballon when the ball is released,
h0 = height of the balloon when the ball is released.
Balloon's acceleration = 4.9 m/s^2
Time to reach height h0 is 2 sec

Therefore, h0 = 1/2 * 4.9 * 2^2 = 9.8 m
v0 = 4.9 * 2 = 9.8 m/s

Fix origin on the ground. Take time = 0 when the ball is released.
Ball's initial h0 = 9.8 m
Acceleration = -g = -9.8 m/s^2
Initial velocity v0 = 9.8 m/s
Velocity at maximum height v = 0
Let maximum height = H

v^2 = v0^2 + 2*(-g)*(H - h0)
Or 0 = 9.8^2 + 2*(-9.8)*(H - 9.8)
Or 0 = 9.8^2 - 2*9.8*(H-9.8)
Dividing by 9.8,
0 = 9.8 - 2*(H - 9.8)
Or 2*(H - 9.8) = 9.8
Or H - 9.8 = 9.8/2
Or H - 9.8 = 4.9
Or H = 9.8 + 4.9
Or H = 14.7 m

Hence option A is correct

Piyush Kumar Behera

Approved Tutor Answer8 Years ago

In the question lets suppose the balloon is released at t=0 sec.

Here initial velocity of the balloon is zero as it is released only.So u=0.

At t=2 sec the velocity of the balloon becomes v=u+at

=>v = 0+4.9 x 2 m/s

=>v = 9.8 m/s

Height moved up by the balloon at t=2 sec is

h= ut+ ½at²

h= 0 x 2 + ½ x 4.9 x 2²

h= 9.8 m

Now let us analyse the next part.

When after 2 seconds the ball is released it gets some initial velocity in the upward direction due to inertia of motion of ball to continue in the direction of motion.

When the ball is released it gets an initial velocity of u=9.8 m/s upwards

So it also goes up to an additional hieght the value of which can be found by using h=v²/2g

h=9.8² / 2 x 9.8=4.9 m

So the maximum hieght attained by the ball is 9.8 m + 4.9 m= 14.7 m

Hence the answer is 14.7m.

I hope it helps!!

Regards

Piyush Kumar Behera