g A cannonball is shot with an initial speed of 62 meters per second at a launch angle of 25 degrees toward a castle wall that is 260 meters away. If the wall is 20 meters tall, how high off the ground will the cannonball hit

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.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

i need a picure

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