physics 203
posted by ash on .
a projectile is thrown with initial speed v0=25.0m/2 at an initial angle with the horizontal ANGLE=45 degrees.
a. find the time when the projectile is at the maximum height.
b. find the position(x,y) at the maximum height.
c. at time T, find the components of the acceleration vector and the velocity vector.
d. Sketch the trajectory of the projectile. On your sketch label the position of the projectile at the time T, draw the velocity vector, and draw the components of the acceleration vector.

a. find vertical and horizontal components of initial velocity.
then, hf=h0+viy*time1/2 g time^2
solve for time.
b. at max height, vy=0
vy=viygt
solve for time t.
c. acceleration is always 9.8m/s
velocity is the vector addition of vy + vix