The projectile launch allows to find the results for questions about the movement are:
a) The initial velocity is: vo = 20 m/s with an angle of θ = 51.4º
b) The maximum height of the ball is: y = 12.5 m
c) The final velocity is: vo = 20 m/s with θ = 308.3º
The launch of projectiles is an application of kinematics where there is no acceleration on the x axis and the y axis is the gravity acceleration.
In the attachment we see a diagram of movement, the x-axis is horizontal and the y-axis is vertical.
They indicate that the range was x = 40 m and the ball was in the air t=3.2s.
a) Let's find the horizontal velocity, as there is no acceleration.
x = v₀ₓ t
v₀ₓ = [tex]\frac{x}{t}[/tex]x / t
let's calculate
v₀ₓ = [tex]\frac{40}{3.2}[/tex]
v₀ₓ = 12.5 m / s
We look for the vertical speed, when the ball reaches the destination its height is zero.
y = [tex]v_o_y[/tex] t - ½ g t²
0 = [tex]v_o_y[/tex] t - ½ g t²
0 = t ( [tex]v_o_y[/tex] - ½ g t)
The solution to this equation is:
t = 0 s Exit time.
[tex]v_o_y[/tex] - ½ g t = 0
[tex]v_o_y[/tex] = ½ g t
It indicates the time in the air that, because it is a scalar, is the same for the two movements.
Let's calculate
[tex]v_o_y[/tex] = ½ 9.8 3.2
[tex]v_o_y[/tex] = 15.68 m / s
We use the Pythagoras' theorem for the initial velocity modulus.
v₀ = [tex]\sqrt{v_{ox}^2 + v_{oy}^2 }[/tex]
v₀ = [tex]\sqrt{12.5^2 + 15.68^2}[/tex]
v₀ = 20 m / s
We use trigonometry for the angle.
tan θ = [tex]\frac{v_o_y}{v_o_x}[/tex]
θ = tam⁻¹ [tex]\frac{v_o_y}{v_o_x}[/tex]
θ = tan⁻¹ [tex]\frac{15.68}{12.5}[/tex]
θ = 51.4º
b) For the maximum height the vertical speed is zero.
[tex]v_y^2 = v_{oy}^2 -2g y[/tex]
0 = [tex]0 = v_{oy}^2 -2 g y[/tex]
y = [tex]\frac{v_{oy}^2}{2g}[/tex]
Let's calculate
y = [tex]\frac{15.68^2}{2 \ 9.8}[/tex]
y = 12.5 m
C) The final velocity of the ball.
Since the x-axis there is no acceleration, the velocity is constant.
vₓ = 12.5 m / s
Let's find the final vertical speed.
[tex]v_y = v_{oy} - g t[/tex]
[tex]v_y[/tex] = 15.68 - 9.8 3.2
[tex]v_y[/tex] = - 15.68 m / s
We are looking for the module.
v = vₓ² + [tex]v_y^2[/tex]
speed gives the same value.
v = 20 m / s
We use trigonometry for the angle.
θ'= tan⁻¹ [tex]\frac{-15.68}{12.5}[/tex]
θ' = -51.66º
This angle measured from the positive side of the x-axis counterclockwise is:
θ = 360 - θ'
θ = 360 - 51.66
θ = 308.3º
In conclusion, using the launch of projectiles we can find the results for questions about the movement are:
a) The initial velocity is: vo = 20 m / s with an angle of θ = 51.4º
b) The maximum height of the ball is: y = 12.5 m
c) The final velocity is: vo = 20 / s with θ= 308.3º
Learn more here: brainly.com/question/15063198
