ball
is
А A
thrown
of
tower with
Take g = 10 m/s²
horizontally from top
velouty of 40 m/s.
a
a
as Find horizontal and vertical displacement after
1, 2, 3, 4, 5 second. then plot the path of 야
mohon of the ball
of the
Answers
Answer:
Let vertical velocity of ball which is thrown above horizontal v
A
=40sin(30
o
)=20ms
−1
,
Let vertical velocity of ball which is thrown below horizontal v
B
=40sin(30
o
)=20ms
−1
,
We know in the projectile when ball comes down after reaching the highest point its magnitude of velocity is same as when it goes upward at same point.
∴ The velocity of ball which is thrown upward is same as velocity of ball which is thrown downward at top of tower when ball is returned at top of tower. So from there both ball takes same time so the time difference is the time taken by ball during upward journey to the top of tower.
At maximum height, Velocity v=0
Using v=u+at ⇒0=20−gt ⇒t=2s
⇒ Total time for upward journey=2t=4s,
∴ Time difference=4s.
Explanation: