An archer shoots an arrow horizontally at a target 11 m away. the arrow is aimed directly at the center of the target, but it hits 56 cm lower. What was the initial speed of the arrow?

Explanation:

Given :

Distance in the vertical  direction, S = 56 cm

                                                               = 0.56 m

Initial velocity in the vertical direction, u = 0 m/s

Acceleration due to gravity, g = 9.81

Therefore, in order to find the time required,

t = 0.33 seconds

Therefore, the initial velocity of the arrow is given by :

     , where S = 11 m and t = 0.33 seconds

Thus the initial velocity of the arrow is 33.33 m/s.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.