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 questions, we can use the kinematic equations for projectile motion. These equations relate the initial velocity, the acceleration due to gravity, time, and the horizontal and vertical displacements of the projectile.
First, let's identify the given values:
Initial horizontal velocity (Vx) = 38.0 m/s
Initial vertical velocity (Vy) = 26.0 m/s
Initial speed (V) = 46.04 m/s
Initial angle (θ) = 34.38°
Time (t) = 2.7 seconds
Acceleration due to gravity (g) = 9.8 m/s^2
1. Maximum height the cannonball goes above the ground:
To find the maximum height, we need to determine the vertical displacement at the highest point of the projectile's trajectory. We can use the following equation:
Vy^2 = V^2 - 2gΔy
First, we need to find the vertical displacement (Δy):
Vy = V * sin(θ)
Δy = Vy * t - 0.5 * g * t^2
Substituting the given values:
Δy = (26.0 m/s) * (2.7 s) - 0.5 * (9.8 m/s^2) * (2.7 s)^2
Calculating this will give you the maximum height the cannonball reaches above the ground.
2. Distance from where it was shot that the cannonball will land:
To find the horizontal displacement, we can use the equation:
Δx = Vx * t
Substituting the given values:
Δx = (38.0 m/s) * (2.7 s)
Calculating this will give you the distance from where it was shot that the cannonball will land.
3. Speed of the cannonball 2.7 seconds after it was shot:
To find the speed, we can use the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
Substituting the given values:
V = √((38.0 m/s)^2 + (26.0 m/s)^2)
Calculating this will give you the speed of the cannonball 2.7 seconds after it was shot.
4. Height above the ground 2.7 seconds after it is shot:
To find the vertical displacement, we can use the equation for Δy:
Δy = Vy * t - 0.5 * g * t^2
Substituting the given values:
Δy = (26.0 m/s) * (2.7 s) - 0.5 * (9.8 m/s^2) * (2.7 s)^2
Calculating this will give you the height above the ground that the cannonball reaches 2.7 seconds after it is shot.