Posted by **jim** on Wednesday, May 18, 2011 at 6:04pm.

Suppose the stone is thrown at an angle of 32.0° below the horizontal from the same building (h = 49.0 m) as in the example above. If it strikes the ground 69.3 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.)

(a) the time of flight s

(b) the initial speed m/s

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

speed m/s

angle ° below the horizontal

- Physics -
**mike**, Friday, May 20, 2016 at 9:38pm
heres two answers i got for the same question.

physics - bobpursley, Friday, May 20, 2016 at 4:49pm

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.

physics - Damon, Friday, May 20, 2016 at 4:55pm

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

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