a cannon fire a bullet into a castle. leave the gun with velocity of 110 with an angle of 20. the bullet reaches highest point at P when it is 1 m above the castle. What is the height of the top of the castle? and waht is the horizontal distamce between the accnon and the castle
To find the height of the top of the castle and the horizontal distance between the cannon and the castle, we can use the equations of projectile motion.
First, let's separate the motion of the bullet into its horizontal and vertical components.
1. Vertical component:
The initial vertical velocity (Viy) can be found using the given initial velocity (110 m/s) and the angle of projection (20 degrees):
Viy = initial velocity * sin(angle)
= 110 m/s * sin(20 degrees)
≈ 37.57 m/s
Now we can find the time taken for the bullet to reach its highest point (P). At the highest point, the vertical velocity will be zero (Vy = 0). We can use the equation:
Vy = Viy - g*t
0 = 37.57 m/s - 9.8 m/s^2 * t
Solving for t, we get:
t ≈ 3.84 seconds
Using this time value, we can find the height of point P above the ground using the equation:
h = Viy*t - (1/2)*g*t^2
= 37.57 m/s * 3.84 s - (1/2)*9.8 m/s^2 * (3.84 s)^2
≈ 72 meters
So, the height of the top of the castle is approximately 72 meters.
2. Horizontal component:
The horizontal velocity (Vix) remains constant throughout the motion. We can find it using the given initial velocity (110 m/s) and the angle of projection (20 degrees):
Vix = initial velocity * cos(angle)
= 110 m/s * cos(20 degrees)
≈ 103.4 m/s
Now, we can find the horizontal distance traveled by the bullet using the equation:
distance = Vix * t
= 103.4 m/s * 3.84 s
≈ 397 meters
So, the horizontal distance between the cannon and the castle is approximately 397 meters.