1. The problem statement, all variables and given/known data A playground is on the flat roof of a city school, 4.9 m above the street below (see figure). The vertical wall of the building is h = 6.00 m high, forming a 1.1-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 was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) _______m/s (b) Find the vertical distance by which the ball clears the wall. ________ m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. _________m 2. Relevant equations Δx=v_x*t, v_y=v_oy+a_y*t, v_y^2=v_oy^2+2a_y*Δy, Δy=1/2(v_oy+v_y)t, Δy=v_oy*t+1/2a_yt^2 3. The attempt at a solution So, pretty much, I tried setting the wall as the high point and then ran from there to find the other parts, but none of the numbers I'm getting work as answer or work together. I'm sorry for my sad attempt at it, but I have no clue what I am doing on this one, its completely different from anything we've done yet and I'm really confused. Any and all help is appriciated :)
Great problem. I did that once with a soccer ball at the start of gym class. Coming outside, I kicked it backwards over my head and onto the roof. No soccer that day as it was the only ball we had. Try drawing a picture of the problem. Also show your work, sometimes in trying to explain it the solution becomes apparent.
Show your workings and we will try to figure where it goes wrong. Writting out all those kinematic formulae is ok but using where appropriate is the most important. Add: Δx=v_x*t, v_y=v_oy+a_y*t, v_y^2=v_oy^2+2a_y*Δy, Δy=1/2(v_oy+v_y)t, Δy=v_oy*t+1/2a_yt^2 You can click icon above for superscript and subscript v_{x} or v^{2}
The key information is the horizontal references. The ball began 24m from the wall, and 2.2 seconds later it was vertically above the wall. That enables you to calculate the horizontal component of the initial velocity. The 53 degree angle [possibly a reference to a 3-4-5 triangle], enables you to get the vertical component then away you go.
Hey, i have the same question to answer. To find the initial velocity of the ball do we only consider the horizontal velocity or must we also find the vertical velocity? If so, how do we find it ?
Because the ball is launched at 53 degrees, if you are given/calculate the horizontal or the Vertical or the actual speed of the ball you can use that angle to calculate any of the others. ie 53 degrees is the angle in a 3-4-5 pythagorean triangle (approx) so if the horizontal speed is 3m/s, the vertical speed is 4 m/s and the actual speed is 5 m/s.
I know this thread is a little old, but I have the same exact question. I can get you started on the problem, I found it a bit difficult at first as well. Remember that the resultant (overall velocity) and its X component and Y component form a right triangle. You are given the angle and you can solve for the X component of the velocity triangle.
i kiked a bal and it go on da rof and i tol da teecher m9 get mai bal m9 or u iz foneμμμμμμμμμμμμμμμμμμμμμμμμμ