Answer:
h = 16.23 m
The cannonball will hit the wall at 16.23m from the ground.
Step-by-step explanation:
Given;
Initial speed v = 62m/s
Angle ∅ = 25°
Horizontal distance d = 260 m
Height of wall y = 20
Resolving the initial speed to vertical and horizontal components;
Horizontal vx = vcos∅ = 62cos25°
Vertical vy = vsin∅ = 62cos25°
The time taken for the cannon ball to reach the wall is;
Time t = horizontal distance/horizontal speed
t = d/vx (since horizontal speed is constant)
t = 260/(62cos25°)
t = 4.627 seconds.
Applying the equation of motion;
The height of the cannonball at time t is;
h = (vy)t - 0.5gt^2
Acceleration due to gravity g = 9.81 m/s
Substituting the given values;
h = 62sin25×4.627 - 0.5×9.81×4.627^2
h = 16.2264134736
h = 16.23 m
The cannonball will hit the wall at 16.23m from the ground.
