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)
vertical height h(t) = 26t - 4.9t^2
max height reached at t = 26/9.8 = 2.653
h(2.653) = 34.49
Judging from parts 3 and 4, they expect the answer to part 1 to be 2.7 seconds.
In that case, v(2.7) = 0, h(2.7) = 34.5m
The range is given by
r = 46^2 sin 68.76° / 9.8 = 201.3m
To find the maximum height the cannonball goes above the ground, we can use the formula for vertical motion:
H = (v₀y²) / (2g)
Where:
H = the maximum height
v₀y = initial vertical velocity
g = acceleration due to gravity (9.8 m/s²)
Plugging in the values:
v₀y = 26.0 m/s
g = 9.8 m/s²
H = (26.0 m/s)² / (2 * 9.8 m/s²)
H = 676.0 m²/s² / (19.6 m/s²)
H = 34.49 m
Therefore, the maximum height the cannonball goes above the ground is approximately 34.49 meters.
To find how far from where it was shot the cannonball will land, we can use the formula for horizontal motion:
R = v₀x * t
Where:
R = the horizontal distance traveled
v₀x = initial horizontal velocity
t = time of flight
Plugging in the values:
v₀x = 38.0 m/s
t = (2 * v₀y) / g = (2 * 26.0 m/s) / 9.8 m/s² = 5.31 s
R = (38.0 m/s) * (5.31 s) = 201.78 m
Therefore, the cannonball will land approximately 201.78 meters from where it was shot.
To find the speed of the cannonball 2.7 seconds after it was shot, we need to find the magnitude of its velocity:
v = √(v_x² + v_y²)
Where:
v = velocity
v_x = horizontal velocity
v_y = vertical velocity
Plugging in the values:
v_x = v₀x = 38.0 m/s
v_y = v₀y - g * t = 26.0 m/s - 9.8 m/s² * 2.7 s = 26.0 m/s - 26.46 m/s = -0.46 m/s (negative indicates downward direction)
v = √((38.0 m/s)² + (-0.46 m/s)²)
v = √(1444.0 m²/s² + 0.2116 m²/s²)
v ≈ √1444.2116 m²/s²
v ≈ 38.04 m/s
Therefore, the speed of the cannonball 2.7 seconds after it was shot is approximately 38.04 m/s.
To find how high above the ground the cannonball is 2.7 seconds after it is shot, we can use the formula for vertical motion:
h = v₀y * t + (1/2) * g * t²
Where:
h = the height
v₀y = initial vertical velocity
g = acceleration due to gravity (9.8 m/s²)
t = time
Plugging in the values:
v₀y = 26.0 m/s
g = 9.8 m/s²
t = 2.7 s
h = (26.0 m/s) * (2.7 s) + (1/2) * (9.8 m/s²) * (2.7 s)²
h = 70.2 m + 35.49 m
h ≈ 105.69 m
Therefore, the cannonball is approximately 105.69 meters above the ground 2.7 seconds after it is shot.