Physics

A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 53.0°; above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

a)Find the speed at which the ball clears the wall.

I calculated this correctly and got 18.13 m/s.

(b) Find the vertical distance by which the ball clears the wall.

I know that dy=vy0*t but I tried using that equation and couldn't get the answer.

I did this and it was incorrect
18.13*sin53*2.2=dy
32.85-6.7=25.45

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.

  1. 0
  2. 0
asked by Lauryn
  1. On part b,
    You need to find how long it takes the ball to go th 24.0m in the x direction. Use that time for the y calculation.

    posted by Quidditch
  2. Sorry, I see that was given.

    posted by Quidditch
  3. Yeah :( thanks for trying though, I tried using the v0sin53 and got 18.13*sin53= 14.48

    then I tried plugging that into the y, to get y=14.48sin53(2.2)-1/2(-9.8)(2.2^2)
    I got 25.44-23.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong

    posted by Lauryn
  4. I calculated the ball was thrown at a speed of 18.12m/s.

    Vx is 24m/2.2s=10.91m/s
    Vx=V(cos(53))
    That gives V as 18.12m/s

    posted by Quidditch
  5. I get 14.47m/s for Vy.

    posted by Quidditch
  6. Using that...
    Y=14.47m/s(2.2s)-(1/2)(9.8m/s)(2.2)^2
    Y=31.834m - 23.716m
    =8.118m
    So it clears the wall by 8.118m - 6.7m
    =1.418m

    posted by Quidditch

Respond to this Question

First Name

Your Response

Similar Questions

  1. college Physics

    A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the
  2. Physics

    A playground is on the flat roof of a city school, 7.00 m above the street below. The vertical wall of the building is 8.00 m high, forming a 1.00 m high railing around the playground. A ball has fallen to the street below, and a
  3. math

    A playground is on the flat roof of a city school, 5.8 m above the street below (see figure). The vertical wall of the building is h = 7.30 m high, to form a 1.5-m-high railing around the playground. A ball has fallen to the
  4. HARD PHYSICS

    A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 7.10 m high, to form a 1.4-m-high railing around the playground. A ball has fallen to the
  5. college Physics I

    A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the
  6. College Physics

    A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, forming a 1.3-m-high railing around the playground. A ball has fallen to the
  7. Physics for Engineers

    A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). The vertical wall of the building is h = 6.70 m high, forming a 1.5-m-high railing around the playground. A ball has fallen to the
  8. math trigonometry

    1. Two straight roads intersect to form an angle of 75". Find the shortest distance from one road to a gas station on the other road 10o0 m from the junction. 2. Two buildings with flat roofs are 60 m apart. From the roof of the
  9. Math

    A school is fencing in a rectangular area for a playground. It plans to enclose the playground using fencing on three sides (One length is a wall) The school has budgeted enough money for 75 ft of fencing material and would like
  10. Math

    A city park planner is deciding between two proposals for a new rectangular playground. One is for a playground with an area of 1930 square feet. A) The second proposal is for a playground that is 11.5 meters long and 19.5 meters

More Similar Questions