Hey guys !
# URGENT #
NEED PROPER AND CORRECT ANSWER WITH FULL STEPS AND EXPLANATIONS !
________________________________________________________
Q. A bird is sitting on a top of the tree 35m high is shot is fire from the distance 70m away from the foot of the tree. In such a direction it just hits the bird. Find the magnitude and direction of velocity with which shot is fired?
________________________________________________________
CLASS 11TH PHYSICS CH :- MOTION IN PLANE !
NO SPAMS PLEASE !
CONTENT QUALITY ANSWER REQUIRED !
Answers
u = 10√14 m/s
θ = 45°
--------------------------------------------------
--------------------------------------------------
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°
◉‿◉
