Two bodies were thrown simultaneously fron the same point, one straight up and the other at an angle of theta=30°to the horizontal. The initial velocity of each body is 20 m/s. Neglecting air resistance, the distance between the two bodies at t= 1.2s later is
Answers
Answer:
The answer will be 24 m
Explanation:
According to the problem one of the bodies a1 straight up means making angle of 90° and the other one a2 went in angle of 30° with the horizontal
The velocity of two bodies is 20 m/s
therefore the relative velocity is
v= √u^2+u^2- 2uxucos(90-θ) [ here θ = 30°]
= √u^2+u^2- 2u^2cos(90-30)
= √u^2+u^2- 2u^2 sin 30
= √2u^2- 2u^2 sin 30
= u√2- 2 sin 30
= u√2-1
= u
As the relative velocity does not change
therefore, the distance covered at 1.2 s
d= u x t= 20 x 1.2
= 24 m
Answer:
22
Explanation:
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:
r
12
=
r
0
(12)
+
v
0
(12)
t+
2
1
w
12
t
2
So,
r
12
=
v
0
(12)
t, (because
w
12
=0 and
r
0
(12)
=0)
or ∣
r
12
∣=∣
v
0
(12)
∣t (1)
But ∣
v
01
∣=∣
v
02
∣=v
0
So, from properties of triangle
v
0(12)
=
v
0
2
+v
0
2
−2v
0
v
0
cos(
2
π
−θ
0
)
Hence, the sought distance ∣
r
12
∣=v
0
2(1−sinθ)
t=22m