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A cannon elevated at 40 degrees is fired at a wall 300 M away on the level ground as shown in figure 3.23. The initial speed of the cannon ball is 3.89 m/s

A How long does it take for the ball to hit the wall?
A 1.3 S
B 3.3 S
C 4.4 S
B at what height h does the ball hit the wall
A 39 m
b 47 m
C 74 m
D 160 m
E 210 m

User Chrisarton
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1 Answer

6 votes

Final answer:

To find the time and height at which an elevated cannonball hits the wall, one must resolve the initial speed into horizontal and vertical components. Time is found by dividing the horizontal distance by horizontal velocity, and the vertical distance is found using the initial vertical velocity, gravitational acceleration, and the time calculated.

Step-by-step explanation:

The projectile motion of a cannonball involves two components: horizontal motion and vertical motion. Since the cannon is elevated, we need to resolve the initial speed into horizontal and vertical components to address each separately.
To find the time it takes for the ball to hit the wall, we analyze the horizontal motion. We use the horizontal velocity (Vx = initial speed × cos(elevation angle)) and the horizontal distance to the wall to calculate time. For the height (h) at which the ball hits the wall, we must first calculate the time of flight using the horizontal component, then use this time to find the vertical position using the formula for the vertical motion.

In summary:

  • Horizontal velocity (Vx): Vx = initial speed × cos(40°) = 3.89 × cos(40°)
  • Time of flight (t): t = distance / Vx = 300 m / Vx
  • Vertical position (y): y = initial speed × sin(40°) × t - (1/2 × g × t^2), where g is the acceleration due to gravity

The exact numerical answers would need calculation with the appropriate formulas.

User Darka
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8.3k points