A stone is thrown up with a velocity which makes an angle theta with the horizontal. After 't' second what will be the magnitude and direction of the resultant velocity?
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a) Magnitude of components
Horizontal component =Ucosθ
vertical component =Usinθ
b) Change with time
i) Ucosθ remains the same.
ii) Usinθ first decreases, becomes zero at highest point A, after which it again increases, till the projectile hits the ground.
c) We have,
vertical motion
U
V
=Usinθ
a
V
=−g
S
V
=H
V
V
=U
V
+a
V
t
⇒V
V
=Usinθ−gt
Horizontal motion
U
H
=Ucosθ
a
H
=0
S
H
=R
V
H
=U
H
+a
H
t
⇒V
H
=Ucosθ
Resultant velocity
V=
V
H
2
+V
V
2
=
(Ucosθ)
2
+(U]sinθ−gt)
2
=
U
2
cos
2
θ+u
2
sin
2
θ+g
2
t
2
−2usinθgt
=
u
2
+g
2
t
2
−2vsinθgt
As sin
2
θ+cos
2
θ=1
V=
U
2
+g
2
t
2
−2Ugtsinθ
solution
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