A cannonball is shot (from ground level) with an initial horizontal velocity of 38.0 m/s and an initial vertical velocity of 26.0 m/s. The initial speed of the cannonball 46.04 m/s The initial angle θ of the cannonball with respect to the ground 34.38°
What is the maximum height the cannonball goes above the ground? (m)
How far from where it was shot will the cannonball land? (m)
What is the speed of the cannonball 2.7 seconds after it was shot? (m/s)
How high above the ground is the cannonball 2.7 seconds after it is shot? (m)
To solve these problems, we can use the equations of motion for projectile motion. The key equations we need are:
1. Vertical displacement equation: Δy = v₀yt + (1/2)gt²
2. Horizontal displacement equation: Δx = v₀xt
3. Vertical velocity equation: vy = v₀y + gt
4. Horizontal velocity equation: vx = v₀x
Where:
- Δy is the vertical displacement (height)
- v₀y is the initial vertical velocity
- t is the time
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- Δx is the horizontal displacement (distance)
- v₀x is the initial horizontal velocity
Let's solve each question step by step:
1. What is the maximum height the cannonball goes above the ground?
The maximum height is achieved when the vertical velocity becomes zero. We can find the time it takes for this to happen using the vertical velocity equation:
vy = v₀y + gt
Substituting the given values:
v₀y = 26.0 m/s
g = -9.8 m/s² (negative because the acceleration due to gravity acts downward)
Setting vy = 0 and solving for t:
0 = 26.0 + (-9.8)t
9.8t = 26.0
t = 2.65 s
Now we can use the vertical displacement equation to find the maximum height (Δy):
Δy = v₀yt + (1/2)gt²
Substituting the values:
v₀y = 26.0 m/s
t = 2.65 s
g = -9.8 m/s²
Δy = 26.0(2.65) + (1/2)(-9.8)(2.65)²
Δy ≈ 34.6 m
Therefore, the maximum height the cannonball goes above the ground is approximately 34.6 m.
2. How far from where it was shot will the cannonball land?
To find the horizontal displacement (Δx), we can use the horizontal displacement equation:
Δx = v₀xt
Substituting the given values:
v₀x = 38.0 m/s
t = 2.65 s
Δx = 38.0(2.65)
Δx ≈ 100.7 m
Therefore, the cannonball will land approximately 100.7 m away from where it was shot.
3. What is the speed of the cannonball 2.7 seconds after it was shot?
The speed of the cannonball is the magnitude of its velocity vector. We can find the velocity components using the equations:
vx = v₀x
vy = v₀y + gt
Substituting the given values:
v₀x = 38.0 m/s
v₀y = 26.0 m/s
g = -9.8 m/s²
t = 2.7 s
vx = 38.0 m/s
vy = 26.0 + (-9.8)(2.7)
vy ≈ 0.6 m/s
The speed of the cannonball can be found using the Pythagorean theorem:
speed = √(vx² + vy²)
speed = √(38.0² + 0.6²)
speed ≈ 38.0 m/s
Therefore, the speed of the cannonball 2.7 seconds after it was shot is approximately 38.0 m/s.
4. How high above the ground is the cannonball 2.7 seconds after it is shot?
To find the height (Δy), we can use the vertical displacement equation:
Δy = v₀yt + (1/2)gt²
Substituting the given values:
v₀y = 26.0 m/s
t = 2.7 s
g = -9.8 m/s²
Δy = 26.0(2.7) + (1/2)(-9.8)(2.7)²
Δy ≈ 40.7 m
Therefore, the cannonball is approximately 40.7 m above the ground 2.7 seconds after it is shot.