The projectile launch relations allow finding the results for the different questions are:
1. Rise time is 2.35 s
2. The maximum height is 27 m
3. The initial velocity is 26 m / s
4. The launch angle is 62.3º
The launch of projectiles is an application of kinematics to the movement of objects near the earth's surface, in this case there is no acceleration on the x-axis and the acceleration is the gravity acceleration on the y-axis.
In the attachment we have a movement diagram of the football ball
1. As the acceleration is constant, the time it takes the body to go up is equal to the time it takes to go down, therefore, as the total time affects, the time it takes to go up is half.
[tex]t_u =\frac{t_{total} }{ 2}[/tex]
t_u = 4.70 / 2
t_u = 2.35 s
2. How high does the ball reach?
At the point of maximum height the vertical velocity is zero
[tex]v_y = v_{oy} - g t\\0= v_{oy} - gt\\v_{oy} = gt[/tex]
[tex]v_{oy}[/tex] = 9.8 2.35
[tex]v_{oy}[/tex] = 23 m / s
Now we can use the equation.
[tex]v_y^2 = v_{oy}^2 - 2 g y\\0 = v_{oy}^2 - 2 g y\\\\y = \frac{v_{oy}^2}{2g}[/tex]
y = [tex]\frac{23^2}{2 \ 9.8}[/tex]
y = 27 m
2. What is the initial velocity?
The initial vertical velocity is 23 m / s
We look for the horizontal speed.
[tex]v_x = \frac{x}{t}[/tex]
They indicate the throw range is 57 m in the time of 4.70 s
vₓ = 57.0 / 4.70
vₓ = 12.1 m / s
To calculate the magnitude we use the Pythagorean Theorem
v₀ = [tex]\sqrt{v_{ox}^2 + v_{oy}^2}[/tex]
v₀ = [tex]\sqrt{12.1^2 + 23^2}[/tex]
v₀ = 26 m / s
3. The launch angle
Let's use trigonometry
tan θ = [tex]\frac{v_{oy}}{v_{ox}}[/tex]
θ = tan⁻¹ [tex]\frac{v_{oy}}{ v_{ox}}[/tex]
θ = tan⁻¹ [tex]\frac{23}{12.1}[/tex]
θ = 62.3º
In conclusion, using the projectile launch relatios we can find the results for the different questions are:
1. The rise time is 2.35 s
2. The maximum height is 27 m
3. The initial velocity is 26 m / s
4. The launch angle is 62.3º
Learn more here: brainly.com/question/10903823
