Share
A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4.8 m to one
Question
A 35 g steel ball is held by a ceiling-mounted electromagnet 3.8 m above the floor. A compressed-air cannon sits on the floor, 4.8 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.4 m above the floor. What was the launch speed of the plastic ball?
in progress
0
Physics
4 years
2021-08-08T17:35:47+00:00
2021-08-08T17:35:47+00:00 1 Answers
412 views
1
Answers ( )
Answer:
Explanation:
Let v is the launch speed of the plastic ball and the angle of projection is θ.
So, in horizontal direction
v Cosθ x t = 4.8 …. (1)
In th evertical direction
1.4 = v Sin θ x t – 0.5 gt² …. (2)
As , v Sin θ x t = 3.8 …. (3) , put in equation (2)
1.4 = 3.8 – 4.9 t²
t = 0.7 s
Put in (1) and (3)
v Cosθ x 0.7 = 4.8
v Cosθ = 6.86
and v Sinθ x 0.7 = 3.8
v Sinθ = 5.43
Now
v = 8.75 m/s