A baseball leaves the bat with a speed of 44m/s and an angle of 30degrees above

Question

A baseball leaves the bat with a speed of 44m/s and an angle of 30degrees above the horizontal. A 5.0 m fence is located at ahorizontal distance of 132m from the point where the ball isstruck. Assume the ball leaves the bat 1.0m above ground level.

A) How long is the ball in the air from the bat to the time it goesover the fence?
B) By how much does the ball clear the fence?

Completely clueless to how to even start off the problem. Usually Idraw a picture to put with the problem but the position of thefence confuses me and where I should draw it. Greatly appreciate astep by step process.

Explanation / Answer

First set your initial values of the relevant physicalquantities:
x1=0       y1=1m      v1x=44*cos(30)m/s       v1y=44*Sin(30)m/s       ax=0       ay=g=9.8m/s^2 . (x1, x1 are the initial values of x and y, v1x, v1y are theinitial x and y components of the velocity, ax and ay are theaccelerations in x and y. we set ax=0 since we are ignoring airresistance for this problem. Also we have that Cos(30)=3/2and Sin(30)=1/2) . Then writte the equations of motion for x and y: .(Lets assume it doesn't hit the ground before it reaches thefence so we write x first. It would be my choice to do part b firstand see if it actually gets that far) x first:      x2=x1+v1x*t+ax*t^2==> x2=v1x*t so we can solve for the time with forx2=132m (the distance to the fence) . .              (132m) /(44*(3/2)m/s) = t then the time for the ball to eithergo over or hit the fence is t=3.46s . Now we do it for the y component: (remember that gravitypoints down so the acceleration is towards negative y) . .               y2=y1+v1y*-g*t^2 ==> y2=1m+44*(1/2)m/s*3.46s - 9.8m/s^2 * (3.46s)^2 .               y2=-40.2m (Negative y2 means the ball hit the ground way before it got tothe fence, so the answer is it didn't clear the fence) . Lets work out when the ball hits the ground: (that it set y2=0and solve the equation for t) . .               0=1+22*t-9.8*t^2    ==>}22±(222 + 4*9.,8*1)}/2*9.8 ==>t=2.3seconds (as expected the quadratic equation has 2 soutions butthe other t is negative so is not physical) .     So we see that the ball hit theground after 2.3 seconds. You can plug this value in the equationfor x and see how far it got but it definitelly doesn't clear thatfence, it only goes a little above 80m before it hits.

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