physics problem.....?

Question

a ball is thrown from the top of a building at an angloe of 58 to the horizontal and with an initial speed of 12.1m/s. The point of realease is 8.5m aobve the ground. find the horizontal range of the ball from the base of the building.

i have 17.41 as the answer but it's wrong....please give me a correct one.

Answer 1

1) Solve for the time it takes for the ball to hit the ground using the formula:
d = Vi*t + 0.5*a*t^2

d = -8.5 m (negative because it is below the starting position)
Vi = 12.1*sin(58 deg) m/s (initial vertical velocity)
a = -9.81 m/s^2 (acceleration due to gravity)

Using the quadratic formula you should get 2.727401 seconds.


2) Solve for the range. Assuming no frictional forces the initial horizontal velocity equals the final horizontal velocity. Thus use the formula: d = V*t

V = 12.1*cos(58 deg) m/s (horizontal velocity)
t = 2.727401 sec

ANS: d = 17.488158 m
2 Significant Figures: d = 17 m


Notes:
- the acceleration due to gravity used was 9.81 m/s^2 and may vary depending on what is the accepted value where you are.
- use the quadratic formula to give you the exact value for time.
- do not round any numbers until the final answer to avoid error.

Answer 2

h=-4.9t^2+12.1*sin58= -4.9t^2+10.261t
v=-9.8t+10.26
t for h max = 9.8/10.26=0.955 s
hmax = 5.33m above the building
Time to the ground from this position( v=0)
13.83=4.9t^2 so t = 1.68s
total time =2.635
so horizontal distance=
d=2.635*12.1*cos 58 =16.896 m

Answer 3

222.8