Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    A ball is projected from the ground towards a vertical wall. It strikes the wall at the highest point of its trajectory, 10 metres above the ground. It bounces horizontally away from the wall, with its speed halved by the impact. It lands on the ground at a distance of 20 metres from its original point of projection.

    Find the ball’s initial speed and angle of projection.

    I know that you will need to use the formulas for the max height and the horizontal range to find the time it takes for the ball to get to the wall, I'm wondering if anyone can verify what value of 't' you get for this since I got 4.472s and I'm not sure if it is correct or not. :/

    Can anyone help?
    Offline

    17
    ReputationRep:
    Can you show your working? Makes it much easier to verify if your value is right or not.
    Offline

    2
    ReputationRep:
    (Original post by As_Dust_Dances_)
    A ball is projected from the ground towards a vertical wall. It strikes the wall at the highest point of its trajectory, 10 metres above the ground. It bounces horizontally away from the wall, with its speed halved by the impact. It lands on the ground at a distance of 20 metres from its original point of projection.

    Find the ball’s initial speed and angle of projection.

    I know that you will need to use the formulas for the max height and the horizontal range to find the time it takes for the ball to get to the wall, I'm wondering if anyone can verify what value of 't' you get for this since I got 4.472s and I'm not sure if it is correct or not. :/

    Can anyone help?
    I do not get the same value for t when it first hits the wall.

    The way I went about it is to call the initial speed x, such that the vertical component of velocity is initially x sin(\theta). You can then write an equation for vertical displacement, keeping in mind that the ball reaches 10 metres at its highest point.

    At the highest point, the vertical velocity will just be zero, so you can the formula v = u + at with v = 0 to eliminate the unknown value x sin(\theta) from the displacement equation, which gives t rather quickly.
    Offline

    3
    ReputationRep:
    (Original post by As_Dust_Dances_)
    A ball is projected from the ground towards a vertical wall. It strikes the wall at the highest point of its trajectory, 10 metres above the ground. It bounces horizontally away from the wall, with its speed halved by the impact. It lands on the ground at a distance of 20 metres from its original point of projection.

    Find the ball’s initial speed and angle of projection.

    I know that you will need to use the formulas for the max height and the horizontal range to find the time it takes for the ball to get to the wall, I'm wondering if anyone can verify what value of 't' you get for this since I got 4.472s and I'm not sure if it is correct or not. :/

    Can anyone help?
    At first I would calculate the vertical component of the speed
    \frac{1}{2}m\cdot v_y^2=m\cdot g\cdot h
    where h=10 m above the ground. From this
    v_y^2=2gh
    Then calculate the time from
    v_y\cdot t-\frac{g}{2}t^2=h
    The horizontal distance for t time
    s_1=v_x\cdot t
    After bounce:
    for the t2 time
    \frac{g}{2}\cdot t_2^2=h
    and with t2
    s_2=\frac{v_x}{2} \cdot t_2
    And you know that
    s_1-s_2=20
    From this you will get v_x
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Brister)
    I do not get the same value for t when it first hits the wall.

    The way I went about it is to call the initial speed x, such that the vertical component of velocity is initially x sin(\theta). You can then write an equation for vertical displacement, keeping in mind that the ball reaches 10 metres at its highest point.

    At the highest point, the vertical velocity will just be zero, so you can the formula v = u + at with v = 0 to eliminate the unknown value x sin(\theta) from the displacement equation, which gives t rather quickly.
    I'm still not quite sure, from this I end up with t = 1.428s? and then I'm not sure where to go from this..
    Offline

    2
    ReputationRep:
    (Original post by As_Dust_Dances_)
    I'm still not quite sure, from this I end up with t = 1.428s? and then I'm not sure where to go from this..
    Next time, please show us some working. Any working will do.

    At least give us some nice labelled diagrams. Mathematicians love diagrams.

    I have attached my working for this problem.
    Attached Images
     
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Brister)
    Next time, please show us some working. Any working will do.

    At least give us some nice labelled diagrams. Mathematicians love diagrams.

    I have attached my working for this problem.
    Thank you, makes a lot more sense now!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.