A child throws a ball with an initial speed of 8 m/s at an angle of 40 degrees above the horizontal. The ball leaves her hand 1 m above the ground. How long is the ball in flight before it hits the ground? What is the magnitude of the ball's velocity just before it hits the ground? At what angle below the horizontal does the ball approach the ground?

Answer 1

it falls
initial speed up = Vi = 8 sin 40
y = yo + Vi t -4.9 t^2
0 = 1 + 8 sin 40 t -4.9 t^2
solve for t, time to hit ground
horizontal velocity u = 8 cos 40 forever
find v at t = time to hit ground
v = Vi -9.8 t (will be negative, down)
tan angle = v/u
magnitude speed = sqrt(u^2+v^2)

Answer 2

When it says magnitude of velocity is that the same as the final velocity before impact? The word "magnitude" threw me off.

Answer 3

magnitude is the size sqrt (u^2+v^2)