1.10. Two bodies were thrown simultaneously from the same point:
one, straight up, and the other, at an angle of 0 = 60° to the horizontal.
The initial velocity of each body is equal to vo = 25 m/s.
Neglecting the air drag, find the distance between the bodies t =
= 1.70 s later.
Answers
Answered by
1
Answer:
24..
Explanation:
Answer
For vertically projected body,
y+1=ut−0.5gt2=16.8 m
x1=0
For the projectile at 30 degrees,
Ux=ucosθ
Uy=usinθ
y2=Uyt−0.5gt2=4.8 m
x2=Uxt
R=(y1−y2)2+(x1−x2)2=24
Hope it helpful for you....
Answered by
1
Answer:
Solution :
The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration: <br>
<br> So,
, (because
and
) <br> or,
(1) <br> But
<br> So, from properties of triangle <br>
<br> Hence, the sought distance <br>
.
Explanation:
hope it helps you!
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