Explanation:
Below is an attachment containing the solution.
Answer:
4 m/s
Explanation:
From the question given above, the following data were obtained:
Maximum range (Rₘₐₓ) = 1.6 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
The initial velocity of the projectile can be obtained as follow:
Rₘₐₓ = u² / g
1.6 = u² / 10
Cross multiply
u² = 1.6 × 10
u² = 16
Take the square root of both side
u = √16
u = 4 m/s
Therefore, the velocity of the projectile is 4 m/s
Answer:
The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
Explanation:
It is given that,
Initial gravitational potential energy of the ball-Earth-floor system is 10 J.
The ball then bounces back up to a height where the gravitational potential energy is 7 J.
Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.
Initial mechanical energy is 10 + 0 = 10 J
Mechanical energy just before the collision is 0 + 10 = 10 J
Final mechanical energy, 7 + 0 = 7 J
Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
The right answer for the question that is being asked and shown above is that: "The object's kinetic energy remains the same." If the net work done on an object is zero, you determine about the object's kinetic energy is that The object's kinetic energy remains the same.
I would google "Earth Science textbook free online" and see what you get.
Good luck!
Hope this helps~!
~{Oh Mrs.Believer}
