a ball is thrown vertically upwards and reaches its peakheight at 4 seconds a. f

Question

a ball is thrown vertically upwards and reaches its peakheight at 4 seconds a. find intial velocity b. find max height c. find time when height is at 1/2 max height, (there aresupose to be 2 times) d. find velocity when height is 1/2 max e. plot 0<T<2 (T=time) any submissions are appreicated a ball is thrown vertically upwards and reaches its peakheight at 4 seconds a. find intial velocity b. find max height c. find time when height is at 1/2 max height, (there aresupose to be 2 times) d. find velocity when height is 1/2 max e. plot 0<T<2 (T=time) any submissions are appreicated

Explanation / Answer

a.    To find the initial velocity, use the equationv=vo+at, where v is the final velocity, vo isthe initial velocity, a is the acceleration, and t is thetime. We know the velocity is 0 after 4 seconds (the ball reaches itspeak height and momentarily stop, after that it begins a downwardmotion). So, we can now plug: v=0 t=4s, and the acceleration (which, in this case, is due togravity) 9.8 ms2 into the equation: 0=vo+(-9.81)*4. Setting down (towardearth) as the negative and up as the positive (for simplicity'ssake it is best to take the initial direction of the particle'smotion as positive and any variables in the opposite direction asnegitive.). Solving for the initial velocity we get 9.81*4=voso vo= 39.24 ms b.    To find the ball's max height we use the equation:x=xo+vo*t+1/2*a*t2: x is thefinial position, xo is the initial position,vo is the initial velocity, a is the acceleration, and tis the time. We use t=4 (which it the time when the ball reaches it's maxheight), a=9.81ms2, vo=39.24, andxo is zero (the problem gives no starting height, so weassume it is at ground level). We solve for the finial hight,remembering that the gravity is negative. x=39.24*4+1/2*-9.81*42 = 78.48m c.    We are trying to solve for the times when the ball isat half its max height. The ball reaches this point twice: once onthe way up and once on the way down. we again use the equation:x=xo+vo*t+1/2*a*t2. This time weknow the max height=78.48 m. First we find half of the max height: 1/2*78.48=39.24 m. So,remembering that down is negaitve, we plug the numbers into theformula: 39.24=39.24*t-1/2*9.81*t2 Move all the variables to one side:1/2*9.81*t2-39.24*t+39.24=0. Then, using the quadratic formula we solve for t: (-(39.24)±((-39.24)2-4*4.905*39.24))(2*4.905) t=1.17s, 6.82s d.    To find the velocities when the ball is at half itsmax height we plug the times into the equationv=vo+a*t 39.24-9.81*1.17=27.75 ms 39.24-9.81*6.82=-27.75 ms e.    Would you tell me what you would like T to be relatedto (i.e. position, velocity, etc.). hope this helps.

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