physics

A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the hill would the ball hit the ground?

I don't understand the following explained by Quittich (sp)?

"Since there is no vertical component, the time to hit the ground is the same as if th ball had been dropped from a height of 50.0m. Calculate how long it takes for the ball to fall 50.0m. Take that time and multiply by the horizontal velocity (given as 10.0 m/s)".

  1. 👍 0
  2. 👎 0
  3. 👁 86
asked by Jon
  1. The time that the ball is in the air just depends on how far it has to fall down to the ground. Since the ball was thrown HORIZONTALLY the initial VERTICAL velocity is 0. So, the ball immediately starts to drop. This happens at the same rate as if you just dropped it from the same height.

    If you quickly rolled a ball off a table and then just as that ball rolled over the edge dropped another ball, they would both hit the ground at the same time because.

    Back to the problem...
    The time to hit the ground is then used to find out how far the ball traveled horizontally. The horizontal velocity is 10.0 m/s (given) and does not change. (Disregard air friction for this). So, the distance from the hill is just (time in the air) * (10.0 m/s).

    Here is the formula to calculate the time in the air:

    distance = (initial distance) + (initial velocity) * time + (0.5)*(acceleration)* (t^2)

    Here, initial distance and initial velocity are both 0. So the equation becomes:
    50.0m = (0.5)*(gravity)*(t^2)
    Solve for t.

    Then use the value you found for t.
    horizontal distance = t * 10.0m/s

    1. 👍 0
    2. 👎 0
  2. t=1.02
    initial velocity=10.0

    I'm confused on what to do but from what you said I used the equation you gave me and plugged the numbers I thought were right and I can get an answer of 31.9

    1. 👍 0
    2. 👎 0
    posted by Jon
  3. that was before I knew you responded

    1. 👍 0
    2. 👎 0
    posted by Jon
  4. OK, but when you get your answers, post them and we'll be sure they look good.

    1. 👍 0
    2. 👎 0
  5. im just not getting anything. Im doing it 2 ways:

    50.0m = (0.5)*(9.8)*(t^2)
    4.9 = t
    4.9*10=49

    or 50.0m = (0.5)*(6.67 x 10^-11)*(t^2)
    3.335 x 10^-11 = t
    3.335 x 10^-11 times 10.0 = I cant even get that

    1. 👍 0
    2. 👎 0
    posted by Jon
  6. OK, here are the steps to go through to solve the time eqution:
    Just to be sure...
    The expression t^2 means t to the second power or t * t.
    Solving...
    50.0m = (0.5)*(9.8 m/s^2)*(t^2)
    50.0m = (4.9 m/s^2)*(t^2)
    dividing both sides by (4.9 m/s^2)
    (50/4.9)s^2 = t^2
    10.2 s^2 = t^2
    taking the square root of both sides
    3.19s = t

    So the horizontal distance is:
    (3.19 s) * (10.0 m/s) = 31.9m

    1. 👍 0
    2. 👎 0
  7. ok. that's what I did it just didn't look right

    1. 👍 0
    2. 👎 0
    posted by Jon
  8. Since there is no initial vertical velocity, gravity pulls the ball down 50 ft in t seconds which derives from h = Vot + gt^2/2.
    SInce Vo = 0, 50 = 9.8t^2/2 making t = 3.19 sec.

    In that 3.19 seconds, the ball travels horizontally a distance of d = 50(3.19) = 159.7m, ignoring air resistance.

    1. 👍 0
    2. 👎 0
    posted by tchrwill
  9. I think the horizontal velocity is 10.0m/s not 50.0m/s.

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math

    A man throws a ball horizontally from the top of a hill 4.9m high. He wants the ball to clear a fence 2.4m high standing on horizontal surface and 8m horizontally away from the point of projection. Find the minimum speed at which

    asked by Kiran on February 13, 2014
  2. Physics

    A man throws a ball horizontally from the top of a hill 4.9m high. He wants the ball to clear a fence 2.4m high, standing on a horizontal surface and 8m horizontally away from the point of projection. Find the minimum speed at

    asked by Kiran on February 13, 2014
  3. physics

    5)A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the hill would the ball hit the ground? 45.0 m 6)A missile launched at a velocity of 30.0 m/s at an angle of 30.0 to the

    asked by Jon on April 1, 2008
  4. physics

    a ball is thrown horizontally from a hill 29.0m high at a velocity of 4.00m/s. find the distance between the base of the hill and the pointwhere the ball hits the ground.

    asked by Anonymous on November 6, 2010
  5. physics

    Two golf balls are thrown with the same speed off the top of a large bridge at the same time. Ball A is thrown straight downward, and Ball B is thrown horizontally off the bridge. Which ball hits the ground first?

    asked by cammi on July 29, 2013
  6. Physics

    Two baseballs are thrown off the top of a building that is 7.24 m high. Both are thrown with initial speed of 63.3 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the

    asked by Gina on September 21, 2010
  7. physics

    A ball is thrown horizontally, with a speed of 20 m/s from the top of a 6.2 m tall hill. How far from the point on the ground directly below the launch point does the ball strike the ground?

    asked by Ryan on October 28, 2010
  8. physics

    A ball is thrown horizontally, with a speed of 20 m/s from the top of a 6.2 m tall hill. How far from the point on the ground directly below the launch point does the ball strike the ground?

    asked by Ryan on October 28, 2010
  9. ap physics

    a ball 1 is thrown horizontally from the top of a building that is 50 meters tall. if another ball 2 is dropped at the same time as the first is released, which ball will hit the ground first? A. ball 1 B. ball 2 c. they land at

    asked by Anonymous on October 31, 2015
  10. physics

    On the surface of the Moon, a ball is thrown horizontally, with a speed of 15 m/s, from the top of a 6 m tall hill. How far from the point on the ground directly below the launch point does the ball strike the ground? (On the

    asked by Karl on September 20, 2011

More Similar Questions