More Physics! projectile motion anyone?

Question

a bullet is fired at 280m/s at an angle of elevation 25 degrees. the bullet can be assumed to be fired from the top of a building that is 75m high. Find:

a) the max height that the bullet reaches ( found = 789.42m)
b) the time of flight
c) the range

Answer 1

Use two sig figs with all answers.

V(y) = 280sin25 = 120

0 = V(y) - at
t(up) = 120/9.8 = 12 sec
x = x0 + vt + 1/2at^2 = 75 + 120*12 - 1/2*9.8*12^2
...= 810 m

t(down) = (2x/a)^0.5 = (2*810/9.8)^0.5
...= 13
t(tot) = 25 sec

y = vt = 280cos25 * 25
...= 6300 m

Answer 2

maximum height= H=
u^2 sin^2(theta)/2g
=280*280 *(sin25)^2divided by 2(10)
time of flight=T=usin(25)/g +[2(75+H)/g]^1/2,where H=the height obtained above.
range=R=ucos(theta)*(time of flight obtained above)
when u will calulate,,you will definitely obtain the answer.
good luck
bond_007----->

Answer 3

H = [U sin 25] ^2 / 2g

H = [280 x sin 25] ^2 / [2x9.8] = 714.42 m

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Vertical displacement = - 75 m
Initial upward velocity = 280 sin 25
Acceleration = - 9.8m/s^2

- 75 = 280 sin 25 t - 0.5 x 9, 8 t^2.

4.9 t^2 - 118.33t -75 =0.

t = 24.77 s, neglecting the negative time value.
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Range = 280 cos 25 x 24.77 = 6285.8 m
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