Question

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.

Answer 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....

Answer 2

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!