1.If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what will be the upward (vertical) velocity at the instant of release? What will be the forward (horizontal) velocity?
How high (above the point of release) will the shot go? What is the time it takes the shot to reach its maximum height?
2.If the shot in the problem above is released from a height of 6 ft and later lands on the ground (height = 0.0 ft),
what was the total time of flight? How far did the shot travel horizontally?
vx=vcos41 = 27.16 ft/s
vy = vsin41 = 23.6 ft/s
dy= (vfy^2-viy^2)/2a) => (0-23.61^2)/[2*(-32)] = 8.71m
vfy = viy + at => 0 = 23.6 + (-32)t => t = 0.7375s
The Attempt at a Solution
for the second part im not sure how to proceed do i use the following equation:
dy= viy^2 + 1/2at^2
6 = 23.6t -16t^t
olving for the quadratic equation i get two value for t
t1 = 1.14848205556934
t2 = 0.32651794443066
By common sense when the height was zero and if i was asked to calculate the total time it would have been 0.7375*2 = 1.475 s. So in part 2 the time should be greater than 1.475s since the height is included so can anyone put me on the correct track please ?