Question

5. A body is projected from height of 60 m with a velocity 10 ms-at angle 30° to
horizontal. The time of flight of the body is [ g =10ms-2)
​

Answer 1

Answer:

h=60 m (downward) (+ve), u

y

=10sin30

o

(initial vertical velocity)

time of light will be time to reach ground i.e. to cover

60 m

↑−ve

displacement

& ↓+ve

Put h=60 m,u→4u=−

2

10

=−5 m/s

9=10 m/s

2

We get

60=−5t+5t

2

or t

2

−t−12=0 or t

2

−2t+3t−12=0 or t(t−4)+3(t−4)=0

⇒t=4 sec

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Answer 2

Explanation:

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