Question

A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height h = 31.0 m above sea level, directed at an angle theta = 47.5° above the horizontal, and with a speed v = 30.6 m/s. Assuming that air friction can be neglected, calculate the horizontal distance d traveled by the projectile.

Answer 1

To find the horizontal distance a projectile travels, calculate the time the projectile is in the air using the vertical velocity component and the initial height, then find the horizontal velocity component and use it with the calculated time to determine the horizontal distance.

Step-by-step explanation:

To calculate the horizontal distance d traveled by the projectile, we need to find the time t the rock is in the air and the horizontal velocity vx. The time t can be found using the vertical motion equation considering the initial height h above sea level:

= vy t - (1/2) g t2, where y is the final vertical position (which is 0 when it hits the sea level), vy is the initial vertical velocity component, and g is the acceleration due to gravity (9.81 m/s2).

Finding the vertical component of the velocity:

vy = v sin(\(θ\))
Therefore,
`vy = 30.6 m/s * sin(47.5°)`

To find the time t, we use the equation setting y = -h because the rock falls 31.0 m below the starting height:

`-31.0 m = (30.6 m/s * sin(47.5°)) t - (1/2) (9.81 m/s2) t2`

After solving the quadratic equation for t, we take the positive root. Then the horizontal distance is found using:

d = vx t, where vx is the horizontal component of the initial velocity.

vx = v cos(\(θ\))
Therefore,
`vx = 30.6 m/s * cos(47.5°)`

Finally, plug in the time t into the equation to find the horizontal distance.