Question

q. 5 and 6.please explain

Answer 1

QUESTION 5

For the balloon,

Initial velocity, u = 14 m/s (upward)

Acceleration, a = -9.8 m/s2 (downward)

Displacement, h = -98 m (downward)

Using,

h = ut + ½ at2

=> -98 = 14t - 1/2 × 9.8 × t2

=> t = 6.12 s

Thus, the packet reaches the ground in 6 s.

Now, using,

v = u + at

=> v = 14 – 9.8 × 6.12 = -45.98 m/s (downward)

This is the velocity with which the packet reaches the ground.


QUESTION 6

If, t1 is the time taken to reach to the maximum height and t2 is the time taken to come to ground from maximum height.

Now, we know that
v = u − gt = 0
t = u/g = 16/9.8 = 1.63 s

Now the body has travelled to the ground in (4 - 1.63) =2 .37 seconds
So using the formula, total height h is

h = 1/2gt^2 = 1/2 × 9.8 ×2.372 = 27.52m

Again the upward distance travelled by the balloon is

h1 = ut − 1/2gt^2 = 16 × 1.63 − 1/2 × 9.8 × 1.632 = 13.06 m

So the height of the balloon is
(27.52 − 13.06) = 14.46 m