Question

The catapult tosses the stone at 50mph and hits a target 100ft away. What was the angle at which the stone leaves the catapult. in degrees

Answer 1

To find the angle at which the stone leaves the catapult, we can use the concept of projectile motion. The angle at which the stone leaves the catapult is approximately 2.80 degrees.

Step-by-step explanation:

To find the angle at which the stone leaves the catapult, we can use the concept of projectile motion. The horizontal speed of the stone is 50 mph and the distance traveled is 100 ft. We can use the formula:

distance = horizontal speed × time

Since we are given the distance and the speed, we can rearrange the formula to solve for time:

time = distance / horizontal speed

Substituting the given values:

time = 100 ft / 50 mph

Converting the time to seconds:

time = (100 ft / 50 mph) × (1 hr / 3600 sec) × (5280 ft / 1 mile)

time = 1.47 seconds

Now we can use the equation for projectile motion:

distance = horizontal speed × time

Solving for the angle:

angle = arctan(distance / (horizontal speed × time))

Substituting the given values:

angle = arctan(100 ft / (50 mph × 1.47 sec))

angle = arctan(100 ft / (50 mph ÷ (1 hr / 3600 sec) × (5280 ft / 1 mile) × 1.47 sec))

angle = arctan(0.049)

angle = 2.80 degrees