On level ground a shell is fired with an initial velocity of 40.0 m/s at 60° above the
horizontal.
-How long does it take to reach its highest point?
-Find its maximum height.
-How far from its firing point does the shell land?
(a) when v=0: 40 sin60° - gt = 0
(b) and (c) see the excellent wikipedia article on Trajectory
To solve this problem, we can use some basic principles of projectile motion. Projectile motion occurs when an object is launched into the air and only influenced by gravity. In this case, the shell is fired at an angle above the horizontal with an initial velocity of 40.0 m/s.
1. How long does it take to reach its highest point?
To find the time it takes for the shell to reach its highest point, we can use the equation for vertical displacement:
𝑑 = 𝑢𝑦𝑖𝑛𝑖𝑡 + 0.5𝑎𝑡^2
Since we are interested in the time it takes to reach the highest point, the final vertical velocity will be zero, and the acceleration will be equal to the acceleration due to gravity, which is -9.8 m/s^2 (taking upward as positive direction).
Vertical equation: d = uy * t + 0.5 * a * t^2
0 = 40 * sin(60°) * t + 0.5 * (-9.8) * t^2
Simplifying the equation:
-4.9t^2 + 20t = 0
Factoring out 't':
t * (-4.9t + 20) = 0
The quadratic equation has two solutions, t = 0 and t = 4.08. Since time cannot be negative, the time it takes to reach the highest point is approximately 4.08 seconds.
2. Find its maximum height.
To find the maximum height reached by the shell, we can use the equation for vertical displacement:
𝑑 = 𝑢𝑦𝑖𝑛𝑖𝑡 + 0.5𝑎𝑡^2
At the highest point, the vertical velocity will be zero, so we can solve for the maximum height.
Vertical equation: d = uy * t + 0.5 * a * t^2
d = 40 * sin(60°) * 4.08 + 0.5 * (-9.8) * (4.08)^2
Simplifying the equation:
d = 20 * 4.08 + 0.5 * (-9.8) * (4.08)^2
The maximum height is approximately 83.58 meters.
3. How far from its firing point does the shell land?
To find the horizontal distance traveled by the shell, we can use the equation for horizontal displacement:
𝑑 = 𝑢𝑥 * 𝑡
Since there is no horizontal acceleration, the initial horizontal velocity remains constant throughout the motion. Thus, we can find the horizontal distance using the initial horizontal velocity and the time it takes for the shell to reach the highest point.
Horizontal equation: d = ux * t
d = 40 * cos(60°) * 4.08
Simplifying the equation:
d = 20 * 4.08
The shell lands approximately 81.6 meters from its firing point.
To solve these questions, we can use the following equations of motion:
- The time taken to reach the highest point (vertical velocity changes from upward to downward) can be found using the equation:
t = (v * sinθ) / g
- The maximum height can be determined using the equation:
H = (v² * sin²θ) / (2 * g)
- The horizontal distance traveled can be calculated using the equation:
D = (v * cosθ) * t
where:
v = initial velocity of the shell (40.0 m/s)
θ = launch angle (60°)
g = acceleration due to gravity (9.8 m/s²)
Let's calculate the answers step-by-step.
1. Time taken to reach the highest point:
t = (v * sinθ) / g
t = (40.0 * sin60°) / 9.8
t ≈ (40.0 * 0.866) / 9.8
t ≈ 3.46 seconds
2. Maximum height:
H = (v² * sin²θ) / (2 * g)
H = (40.0² * sin²60°) / (2 * 9.8)
H ≈ (1600 * 0.75) / 19.6
H ≈ 61.22 meters
3. Horizontal distance traveled:
D = (v * cosθ) * t
D = (40.0 * cos60°) * 3.46
D ≈ (40.0 * 0.5) * 3.46
D ≈ 69.2 meters
Therefore, the answers are:
- The shell takes approximately 3.46 seconds to reach its highest point.
- The maximum height reached by the shell is approximately 61.22 meters.
- The shell lands approximately 69.2 meters from its firing point on level ground.