Question

A bird is sitting on top of a tree 35 m high. A shot is fired from a distance of 70 m from the foot of the tree such that it just hits the bird. Calculate the
magnitude and direction in which the shot is fired.


Please answer this asap.
I have my exam tomorrow.
I'll mark the best answer as brainliest!

Answer 1

this is form a right angle triangle

so,

35^2 + 70^2 = x^2

1225+4900=x^2

6125=x^2

x=78.26m

Answer 2

Answer:

GIVEN —

Height of tree = 35 m

Distance of gunman from foot of tree = 70 m

g = 10 m/s² (for JEE & NEET)

 

To find—

Velocity of bullet = ?

Direction of Velocity = ?

 

Let the velocity of bullet be u m/s

 

The velocity of horizontal component is ucosθ

 

The velocity of vertical component is usinθ

 

Now,

We will apply work energy theoram along vertical component

 

Initial Energy = Final Energy

0 + 1/2 m(usinθ)² = mg×35 + 0

By solving

usinθ = 10√7 m/s

 

Using law of motion along vertical

V = u + gt

⇒0 = 10√7 + 10t

⇒t = √7 sec

 

Same time will be taken to complete horizontal distance

ucosθ = 70/√7

⇒ucosθ = 10√7 m/s

 

HENCE u will be √(10√7 + 10√7)

 

u = 10√14

 

Therefore the velocity of bullet = 10√14 m/s

 

 

Direction means the angle of projection (θ)

 

Looking the attachment

 

tanθ =p/b

⇒tanθ = 10√7/10√7

⇒tanθ = 1

⇒θ = 45°

 

Hence, the angle of projection is 45°