Suppose the stone is thrown at an angle of 40.0° below the horizontal from the same building (h = 42.0m) as in the example above. If it strikes the ground 36.7 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

Question

A)time of flight

b) initial speed

c)the speed and angle of the velocity vector with respect to the horizontal at impact.

if needed i can show you my work so far.

Answer 1

break the initial velocity into two componnets;
horizontal: Vo*cos40
vertical: -Vo*sin40 - 9.8 t

a. hf=-42=-vo*sin40*t-1/2 9.8 t^2
d=36.7=vo*cos40t
solve for vo*t in the second in terms of t. Put that in the first equaiton.
b. knowing t, use the second equation to solve for vo.

Answer 2

42 = Vo t + 4.9 t^2
Vo = speed sin 40

36.7 = u t
u = speed cos 40
so
speed * t = 36.7/cos 40
so
speed * sin 40 * t = 36.7 tan 40
Vo t = 36.7 tan 40
so back to
42 = Vo t + 4.9 t^2
now
42 = 36.7 tan 40 + 4.9 t^2
solve for t, time in the air, and go back and get the rest