Verified answer

In order to solve this, and all projectile motion questions, you need to break it apart into x and y components.

In the x direction, you have no acceleration and constant velocity because we are disregarding wind resistance, so you have v=d/t and the x component of d is what we're looking for.

And in the y direction, you have acceleration due to gravity. In order to find the x distance that the question is asking, you need to first find t in the y direction as we are given more info that way and we need it as it is the connector between the two equations because time doesn't change. For the y direction you will have one of the 5 acceleration formulas to manipulate to solve for t.

X: dx=???

. vx= cos27 x 49 m/s..... (horizontal component of the initial velocity)

. t=? (will find using y component)

Y: Vyi=sin27 x 49 m/s... (vertical component of the initial velocity)

. Vyf=-Vyi ....(how ever hard you throw something upward, it will come down with the same velocity)

. ay= -9.81m/s^2

. t=?

ay=(Vyf-Vyi)/t ----> rearrange for time

t= (Vyf-Vyi)/ay ----> plug and solve

t= (-22.25-22.25)/-9.81

t= 4.54s

Plug t into your X component equation v=d/t or d=vt

d=(49 x cos27)(4.54)

d=43.66 x 4.54

d=198.21

d=198m

Hope that helped! :)