Answer:
32.46m/s
Explanation:
Hello,
To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

Where
Vf = final speed
Vo = Initial speed
=7.3m/S
A = g=acceleration
=9.81m/s^2
X = displacement
=51m}
solving for Vf

the speed with the ball hits the ground is 32.46m/s
In 2011 NASA researchers identified features on Mars known as Recuring Slope Lineae ( or RSL ). RSL are relatively dark and narrow features.They are thought to be a signs of salty liquid water on Mars. One hypothesis for RSL formation is that they form when underground bodies of salty water leak into the surface.
Answer: Recuring Slope Lineae.
Explanation:
At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.
g= acceleration due to gravity
Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons


Now by putting the values

F= 600 N