A cannonball is shot from a cannon on a cliff at a height of 18.9 m at a ship on the ocean. The cannonball is shot with an initial velocity of 15.1 m/s at an angle of 24.7 degrees above the horizontal.
#1- What is the total maximum height of the cannonball (the height with respect to the ocean)?
#2- What is the total time in the air for the cannonball?
#3- What is the total range of the cannonball?
First, resolve the velocity into the horizontal and vertical components,
v0=15.1 m/s
θ=24.7°
vx=v0 cosθ
vy=v0 sinθ
The horizontal velocity will stay constant, while the vertical velocity will be subject to the influence of gravity.
The height above the ocean at time t is given by the equation:
y(t)=y0+vy*t-(1/2)gt²
where
g=9.8 m/s², acc. due to gravity
y0=18.9 m = height of cliff
The vertical velocity at time t is given by
v(t)=vy-g*t
a. The cannonball reaches the highest point at time t when v(t)=0
v(t)=vy-g*t=0
tmax=vy/g
Maximum height
y(tmax)=y0+vy*tmax-(1/2)g*tmax²
b. When the cannonball hits the sea, y(t)=0. The time t1 when this happens is
y(t1)=y0+vy*t1-(1/2)gt1²=0
Solve for t1.
c. Since the horizontal velocity is constant at vx, the total range is
Distance = vx*t1
Post if you have questions or if you wish to have your answers checked.
To find the answers to these questions, we can use kinematic equations and the principles of projectile motion. Here's how:
#1 - To find the maximum height, we need to know the vertical component of the initial velocity. We can calculate it using the formula:
Vy = V * sin(θ), where Vy is the vertical component, V is the initial velocity, and θ is the launch angle.
Substituting the given values:
Vy = 15.1 m/s * sin(24.7°)
Vy ≈ 6.27 m/s
Next, we can use the kinematic equation for vertical motion to find the time taken to reach the maximum height. The equation is:
Vy = Voy + at, where Voy is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is time.
Since the cannonball has no initial vertical velocity, we have:
6.27 m/s = 0 + (-9.8 m/s^2) * t
Solving for t:
t ≈ 0.639 seconds
Now, we can use the formula for vertical displacement to find the maximum height:
Δy = voy * t + (1/2) * a * t^2
Δy = 0 * t + (1/2) * (-9.8 m/s^2) * (0.639 s)^2
Δy ≈ -3.14 meters (negative sign indicates displacement above the launch point)
The total maximum height of the cannonball above the ocean is 18.9 m - 3.14 m = 15.76 m.
#2 - The total time in the air can be calculated using the formula for vertical motion:
Δy = voy * t + (1/2) * a * t^2
Since the cannonball moves vertically upward and then falls back down, the total displacement is zero. So we can write:
0 = 6.27 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Solving this quadratic equation will give us the total time in the air. Using the quadratic formula, we get:
t ≈ 1.29 seconds
#3 - The total range of the cannonball is the horizontal distance it travels. We can calculate it using the formula:
R = Vx * t, where Vx is the horizontal component of the initial velocity, and t is the total time in the air.
To find Vx, we use:
Vx = V * cos(θ), where Vx is the horizontal component, V is the initial velocity, and θ is the launch angle.
Substituting the given values:
Vx = 15.1 m/s * cos(24.7°)
Vx ≈ 13.63 m/s
Finally, we can calculate the range:
R = 13.63 m/s * 1.29 s
R ≈ 17.57 meters
Therefore, the answers to the questions are:
#1 - The total maximum height of the cannonball above the ocean is approximately 15.76 meters.
#2 - The total time in the air for the cannonball is approximately 1.29 seconds.
#3 - The total range of the cannonball is approximately 17.57 meters.