A ball is thrown at an angle of 53 degrees above the (check my answers please)?

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 = 15 m/s

Time = 2.0s, x = 18m, y = 4.4m, Vx= 9m/s, Vy =-7.6m/s and V = 15 m/s

Time = 4.0s, x = 9m, y = -15.2m, Vx= 9m/s, Vy =-27.2m/s and V = 15 m/s



thanks

Answer 1

Recheck T=4. What happened? Did the ball bounce back towards you?

Answer 2

the speed vector has aspects Vx= 15 cos fifty 3 and Vy 15 sin53 Vx=9.03 m/s and is consistent Vy (initial) =11.98m/s So the equation of the vertical flow is Vy =11.ninety 8-9.8t and Y = -9.8/2 t^2 +11.98t (reference the roof of the buiding) V(1s) = sqrt(9.03^2 +(11.ninety 8-9.8)^2)) = 9.29 m/s Y(1s) = 7.08 m above the roof and 9.03m distant horizontally V(2s) = sqrt(9.03^2 +(11.ninety 8-19.6)^2)=11.80 2 m/s(pointing downwards) Y(2s) = 4.36m above the roof and 18.06m distant horizontally you are able to now do it for 4 s

Answer 3

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I hope it helps you to start to move your own brains at last.