A helium balloon leaves the ground at a point A and rises vertically at uniform speed. At the end of

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04 Oct 2019 Heights and Distance

A helium balloon leaves the ground at a point A and rises vertically at uniform speed. At the end of 5 minutes, Raju finds the angular elevation of the balloon as 30°. If the point at which Raju is standing is 550 m away from point A, what is the speed of the balloon in m/s?

OPtion

1) 2.67
2) 2.17
3) 1.67
4) 1.067
5) 1.58
6) 2.058
7) 1.085
8) 2.076
9) 1.058
10) None of these

Solution

Let C be the position of Raju. Let A be the position at which balloon leaves the Ground and B be the position of the balloon after 5 minutes.

Given that CA = 550 m, angle BCA = 30°

tan 30° = 1/(3^1/2) = BA/CA
BA = CA/(√3) = 550/√3 m

time taken = 5 min = 5 × 60 = 300 seconds

Speed = Distance/Time
=> (550/√3)/300
= 1.058 meter/second

Option 9)

Correct Option: 9 Best Solution (3)

G.S.Kantharao   4 years AGO

Tan 30°=1/√3=x/550
x=550/√3
Speed per second =550/300√3=1.058m/s
Right option is 9👈

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SAYAN CHATTERJEE   4 years AGO

The elevation is tan30*550= 317.54
So, velocity is 317.54/300
= 1.058 m/s.
Option 9.

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Elizabeth Pothen   4 years AGO

Height upto which ballon rises = tan 30 * 550 = 317.5426 m
Speed = 317.5426/ (60*5) = 1.058 m/s

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