Question

A hot-air balloon is rising vertically with a constant velocity of 10 m/s. A object is dropped from the balloon when it is 60 m above the ground. The angle of elevation of the sun is 45o. When the falling object is 33.75 m from the ground, the rate at which the shadow of the object is travelling along the ground is (Take g = 10 m/s2)

1. 15 m/s

2. 20 m/s

3. 25 m/s

4, 30 m/s

Answer 1

have a look at the figure below...



 

 

we know that

s = ut + (1/2)at2

so, here the displacement as a function of time will be

s(t) = 60 - 4.9t2

now,

given s = 33.75 m

so,

33.75 = 60 - 4.9t2

thus,

t = [26.25/4.9]1/2

or

t = 2.32 s

now,

tan45 = 1 = s(t) / x

so,

x = s(t) = 60 - 4.9t2

differentiating with respect to time 

dx/dt = s'(t) = -9.8t

or

dx/dt = -9.8.(2.32)

thus, the rate at which the shadow move on the ground will be

v = dx/dt = -22.736 m/s

Was this answer helpful0



Messi answered this
1 unhelpful votes in Physics, Class XII-Science

20

Was this answer helpful0

View More