A projectile of mass 0.998 kg is shot from a cannon. The end of the cannon’s barrel is at height 7 m, as shown in the figure. The initial velocity of the projectile is 11 m/s.The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
To find the maximum height (∆y) and horizontal distance (∆x) of the projectile, we can use the principles of projectile motion.
Step 1: Break down the initial velocity into its horizontal and vertical components.
Given the initial velocity (v) of 11 m/s, we can find its horizontal component (v_x) and vertical component (v_y).
v_x = v * cos(theta)
v_y = v * sin(theta)
Step 2: Determine the time taken to reach maximum height.
To find the time taken to reach the maximum height, we can apply the kinematic equation for vertical motion:
∆y = (v_y * t) + (0.5 * g * t^2)
Since the projectile goes up to the maximum height and then falls back down, the time taken to reach the maximum height is half of the total flight time.
t_max = t_total / 2
Step 3: Calculate the maximum height (∆y).
Using the time taken to reach the maximum height, we can substitute it into the kinematic equation to solve for ∆y.
∆y = (v_y * t_max) - (0.5 * g * t_max^2)
Step 4: Calculate the total flight time.
Using the vertical component of the initial velocity, we can find the total flight time by applying the kinematic equation for vertical motion:
∆y = (v_y * t_total) - (0.5 * g * t_total^2)
Since the final vertical displacement (∆y) is zero (the projectile lands at the same height as it was launched), we can solve for t_total.
Step 5: Calculate the horizontal distance (∆x).
Using the total flight time, we can find the horizontal distance (∆x) traveled by the projectile.
∆x = v_x * t_total
By following these steps and plugging in the given values for velocity (v), height (h), and mass (m), we can calculate the maximum height (∆y) and horizontal distance (∆x) of the projectile.