Answer:
15.7 m/s
Explanation:
The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:
where
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration
t is the time
If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:
Answer:
28√3 m
Explanation:
A = vertex where receiver is placed
S = focus
Bp = r = radius of the outside edge
Bc = 2r = diameter
The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.
Answer:
Explanation:
Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.
Horizontal distance = 13 m
Vertical distance = 57 cm
Lets take time to cover 57 cm distance in vertical direction is t.
We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.
Here
S= 57 cm
t=0.34 s
Now in the horizontal direction
Here x=13 m
t= 0.34 s
So
So the initial speed of ball is 38.13 m/s.
