Calculating Distance, Velocity and Gravity

  • Thread starter Thread starter Kdawg
  • Start date Start date
  • Tags Tags
    Gravity Velocity
Click For Summary
SUMMARY

The discussion focuses on calculating the effects of gravity on projectile motion using specific examples involving an archer and a beach ball. For the archer, the arrow must be aimed 0.975 m above the bull's-eye to compensate for gravity when shot horizontally at 95.0 m/s from a distance of 42.0 m. In the beach ball scenario, the pier height is determined to be 0.264 m above the water, calculated using the ball's horizontal velocity of 1.30 m/s and the distance it traveled (0.73 m) before hitting the water.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion.
  • Familiarity with the equations of motion, including d = v*t and d = 1/2 * g * t^2.
  • Knowledge of gravitational acceleration, specifically 9.8 m/s².
  • Ability to perform algebraic manipulations to solve for time and distance.
NEXT STEPS
  • Study the derivation and application of the equations of motion in different contexts.
  • Explore the effects of varying initial velocities on projectile trajectories.
  • Investigate real-world applications of projectile motion in sports and engineering.
  • Learn about air resistance and its impact on projectile motion calculations.
USEFUL FOR

Students and educators in physics, engineers working on projectile design, and anyone interested in understanding the principles of motion and gravity in practical scenarios.

Kdawg
Messages
34
Reaction score
0
An archer stands 42.0 m from the target. If the arrow is shot horizontally with a velocity of 95.0 m/s, how far above the bull's-eye must he aim to compensate for gravity pulling his arrow downward?

A beach ball, moving with a speed of +1.30 m/s rolls off a pier and hits the water 0.73 m from the end of the pier. How high above the water is the pier?

I need some help with these.
 
Physics news on Phys.org
What have you got so far?
 


To calculate the distance, velocity, and gravity involved in these situations, we can use the equations of motion. For the archer, we can use the equation d = v*t, where d is the distance, v is the velocity, and t is the time. In this case, we are given the distance (42.0 m) and velocity (95.0 m/s), so we can solve for the time it takes for the arrow to reach the target. The equation becomes t = d/v, so t = 42.0 m / 95.0 m/s = 0.442 seconds.

Now, we can use the equation d = 1/2 * g * t^2, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time. We know the time (0.442 seconds) and we want to find the distance above the bull's-eye, so we rearrange the equation to solve for d, giving us d = 1/2 * g * t^2 = 1/2 * 9.8 m/s^2 * (0.442 s)^2 = 0.975 m. This is the distance the arrow will drop due to gravity, so the archer should aim 0.975 m above the bull's-eye to compensate for this drop.

For the beach ball scenario, we can use the same equations. We are given the velocity (1.30 m/s) and the distance traveled (0.73 m), so we can solve for the time using the equation t = d/v. This gives us t = 0.73 m / 1.30 m/s = 0.562 seconds.

Next, we can use the equation d = 1/2 * g * t^2 to find the height of the pier. We know the time (0.562 seconds) and we want to find the height, so we rearrange the equation to solve for h, giving us h = d / (1/2 * g * t^2) = 0.73 m / (1/2 * 9.8 m/s^2 * (0.562 s)^2) = 0.73 m / 2.76 = 0.264 m. This is the height of the pier above the water.

In summary, by using the equations of motion, we can calculate the distance
 

Similar threads

Replies
5
Views
2K
Replies
9
Views
6K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
6
Views
4K
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
18
Views
2K