check my answers (velocity and time problem)?

Question

A ball is thrown at an angle of 53 degrees above the horizontal off the roof of a building with an initial speed of 15 m/s. Find the position and velocity of the ball at 1.0s, 2.0s and 4.0s after it was thrown (ignore air resistance) (x- components are horizontal y components are vertical)

What is the x, y, Vx, Vy and V for the ball at 1.0s, 2.0s and 4.0s

I used equation V=at+Vo to solve for V and X-Xo=.5(V+Vo)t to solve for x.

I got 9m/s for Vox, 12m/s for Voy, ax = -9.8m/s^2 and ax= 0m/s^2

Time = 1.0s, x = 9m, y = 7.1m, Vx= 9m/s, Vy =2.2m/s and V = 9.26 m/s

Time = 2.0s, x = 18m, y = 4.4m, Vx= 9m/s, Vy =-7.6m/s and V = 4.82 m/s

Time = 4.0s, x = 36m, y = -15.2m, Vx= 9m/s, Vy =-27.2m/s and V = -25.66 m/s

Someone else told me the velocitys were wrong can someone show me what to do if I did them wrong, thanks

Answer 1

I dont have time to answer but try http://wikipedia.org/projectile
It gives you formulas

Answer 2

You answer is generally correct, except
three of your values are screwed up a little bit.
It's no big deal.



Vox = V cos θ = 9.03 m/s
Voy = V sin θ = 11.98 m/s
g = 9.80 m/s²

X = Xo + Vox T = 9.03 m/s T
Y = Yo + Vox T + ½g T² = 11.98 m/s T - 4.90 m/s² T²
V = √(Vox² + Voy²)

t X Y Vx Vy V
0 0 0 9.03 11.98 15
1 9.03 7.02 9.03 2.18 9.03
2 18.06 4.36 9.03 -7.62 [[ 11.82 ]]
4 36.12 [[ -30.48 ]] 9.03 -27.22 [[ 28.68 ]]

Answer 3

x1=v t cos53=15 1 0.6 =9m
y1=gt^2/2+ v t sin53=10x1/2+15 1 0.8=5+12=17m,
,

Answer 4

it is the comparable question that Erin asked approximately 2 to 3 hours in the past. See my answer there and consider out to apply the help I surely have given to do it your self until now finding at any precise answer here. (be conscious that I surely have seen somewhat countless incorrect solutions published in this internet site.)