A particle is projected from origin in XY plane at angle 37° with X-axis with a speed of 10m/s. Its acceleration is (-5i+10j)m/s^2. Find the speed of the particle where its x-coordinate is zero
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11th
Physics
Motion in a Plane
Projectile Motion
A particle is projected fro...
PHYSICS
A particle is projected from ground with velocity 40 m/s at 60
∘
with horizontal. Find speed of particle when its velocity is making 45
∘
with horizontal. Also find the times (s) when it happens. (g=10m/s
2
)
December 20, 2019avatar
Vyshnavi Chinnu
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ANSWER
Given : u=40m/s a
y
=−g=−10m/s
2
∴ u
x
=ucos60
o
=20m/s and u
y
=usin60
o
=34.64m/s
x direction : a
x
=0 ⟹V
x
=u
x
=20m/s
Also
V
x
V
y
=tan45
o
=1 ⟹V
y
=V
x
=20m/s
y direction : V
y
=u
y
+a
y
t
Upward motion : 20=34.64−10t
1
⟹t
1
=1.464 s
Downward motion : −20=34.64−10t
2
⟹t
2
=5.464 s
solution
The speed of the particle where its x-coordinate is zero is 8m/s.
Step by step calculation:
Given:
- Angle = 37°
- Speed (u) = 10m/s
- Acceleration = (-5i+10j)m/s²
- x co-ordinate (S) = 0
In x-direction:
Using the equation = u x cos 37 x t =1/2 x at²
t = u x 2/5 x cos37
t = u x 2/5 x 4/5
t = 8u/25
Using the equation→ vₓ = uₓ + at
vₓ = u x cos37 + at
vₓ = 10 x 4/5 -5 x 8u/25
vₓ = 8 -16
vₓ = -8m/s
The speed of the particle where its x-coordinate is zero is 8m/s.
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