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
[tex]\frac {Vf^{2}-Vo^2}{2.a} =X[/tex]
Where
Vf = final speed
Vo = Initial speed =7.3m/S
A = g=acceleration =9.81m/s^2
X = displacement =51m}
solving for Vf
[tex]Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s[/tex]
the speed with the ball hits the ground is 32.46m/s
Answer:
The speed will be 33 m/s.
Explanation:
The equations for height and velocity of the ball are as follows:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity at time "t".
Let´s place the origin of the frame of reference on the ground and consider the upward direction as positive.
First, let´s find the time it takes the ball to reach the ground:
y = y0 + v0 · t + 1/2 · g · t²
When the ball hits the ground, y = 0. Then:
0 = 51 m - 7.3 m/s · t - 1/2 · 9.8 m/s² + t²
Solving the quadratic equation:
t = 2.6 s
Now, with this time, we can calculate the velocity at the moment when the ball hits the ground:
v = v0 + g · t
v = -7.3 m/s - 9.8 m/s² · 2.6 s
v = -33 m/s
The speed will be 33 m/s.
