Respuesta :
Answer:
Using the range formula S = V^2 * sin 2 theta / g
So the cannonball should travel
S = 35^2 * sin 90 / 9.8 = 125 m
Then the cannon should be placed 100 m from the wall
H = Vy t - 1/2 g t^2 height of cannonball after time t
Vy = 35 * sin 45 = 24.7 m/s vertical speed of cannonball
Vx = 35 * cos 45 = 24.7 m/s
t = 100 m / 24,7 m/s = 4.04 sec time for cannonball to reach wall
H = Vy t - 1/2 g t^2 height of cannonball after time t
H = 24.7 * 4.04 - 1/2 * 9.8 * 4.04^2 = 19.8 m
19.8 m - 15 m = 4.8 m cannonball clears wall by 4.8 m
The projectile launch ratios allow to find the results on the cannonball launch questions are:
a) The canyon is at x = 100 m from the wall.
b) The bullet hits the wall at a height of 4.8 m above the height of the wall.
Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is gravity acceleration.
A) Indicate that the initial speed of the bullet is v = 35 m/s with an angle θ=45º, which is the distance from the wall. The target is 25 m behind the wall, in the attachment we have a scheme of the system.
Let's find the range that is the horizontal distance for which the height is zero.
[tex]R = \frac{v_p^2 \ sin \ 2\theta}{g}[/tex]
Let's calculate.
R = [tex]\frac{35^2 \ sin ( 2 45(}{9.8}[/tex]
R = 125 m
Therefore the canonl must be 100 m from the wall.
B) The height of the wall is y = 15m, the blah manages to pass it and by how much.
We use trigonometry to find the components of velocity.
v₀ₓ = v₀ cos θ
[tex]v_{oy}[/tex] = v₀ sin θ
v₀ₓ = 34 cos 45 = 24.7 m / s
[tex]v_{oy}[/tex] = vo sin 45 = 24.7 m / s
Let's find the time it takes for the cannonball to travel the x = 100 m
[tex]v_x = \frac{x}{t} \\ t = \frac{x}{v_x}[/tex]
t = [tex]\frac{100}{35 \ cos 45 }[/tex]
t = 4.04 s
Let's find the height for this time.
y = [tex]v_[oy}[/tex] t - ½ g t²
Let's calculate
y = 24.7 4.04 - ½ 9.8 4.04²
y = 19.8 m
This height is greater than the height of the wall h = 15m, therefore the cannoball passes the wall, by a difference in height.
Δy = y -h
Δy = 19.8-15
Δy = 4.8 m
In conclusion, using the projectile launch ratios we can find the results on the cannonball launch questions are:
a) The canyon is at x = 100 m from the wall.
b) The cannonball hits the wall at a height of 4.8m above the height of the wall.
Learn more here: brainly.com/question/10903823
