A ball is projected with speed 20 m/s at an angle of 60° to the horizontal. Find also the height of the ball
above the level of the point of projection when it has travelled a horizontal distance of 10 m.
(g = 10 m/s)
Answers
Given : u=20
3
m/s θ=60
o
Horizontal component of velocity initially u
x
=u cos60=20
3
×
2
1
=10
3
m/s
Since, there is no acceleration in horizontal direction. So horizontal component of velocity remains the same all the time.
Final horizontal component of velocity v
x
=u
x
=10
3
m/s
Final velocity makes an angle 30
o
with horizontal.
So, vertical component of final velocity v
y
=u
x
tan30=10
3
×
3
1
=10
Initial vertical comonent of velocity u
y
=usin60=20
3
×
2
3
=30 m/s
Using v
y
=u
y
−gt
∴ 10=30−10t
⟹ t=2 s
Answer:
answer is 2
Explanation:
Since, there is no acceleration in horizontal direction. So horizontal component of velocity remains the same all the time.
Final horizontal component of velocity v
x
=u
x
=10
3
m/s
