A particle is projected at a speed of 4 ms at an angle of 15° on a horizontal surface. What is the speed with which it hits the surface on landing? (in ms"') RESPONSE AREA?
Answers
Answer:
Correct option is
C
15.4 m
∴v=v
x
i
^
+v
y
j
^
v=i×u
x
cosθ
i
^
+u
x
×cosθ×tan(θ/2)
j
^
v
2
=u
2
cos
2
θ+u
2
cos
2
θ×tan
2
θ/2
v
2
=u
2
cos
2
θ(1+tan
2
θ/2)
v
2
=u
2
cos
2
θ×sec
2
θ/2
∴ Radius of curvature =
gcosθ/2
u
2
cos
2
θ×sec
2
θ/2
=
4×3×10×
3
400×1×4×2
3
3
80
Explanation:
solution: Correct option is
D
40
3
m/s
Time of flight =
g
2usinθ
=
10
2×80×sin30
o
=8
As particle during journey at t=2sec and at t=6sec will be at same height and hence vertical displacement is zero.
Horizontal displacement =(ucosθ)×(6−2)
=80×
2
3
×4
=160
3
m
∴ Average velocity =
4
160
3
=40
3
m/sec
solution
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