Verified answer

1) Draw the projectile motion.

2) Notice that at the highest point, vertical velocity Vy = 0. So if the question says that its speed when it reaches its highest point is 57 m/s, that means Vx = 57.

3) Vx does not change in projectile motion because there is no force affecting it. On the other hand, gravity affect Vy, that's why Vy changed over time. So we can conclude that Vx = 57 initially and always.

4) Since we now have Vx and the initial velocity(Vi) 88m/s we can use a cosine relationship to relate the two. cos(θ) = Vx/Vi

so θ = arccos(Vx/Vi), plug in given values Vx and Vi and you get the answer.

Magnitude of the velocity = (Vertical velocity^2 + Horizontal velocity^2)^0.5

Tangent of the angle, above the horizontal = Vertical velocity ÷ Horizontal velocity

As the stone moves upward, its vertical velocity decreases 9.8 m/s each second. At the highest point the vertical velocity is 0 m/s. The horizontal velocity remains constant during the entire trip.

So horizontal velocity = 57 m/s

88 = (Vertical velocity^2 + 57^2)^0.5

88^2 = Vertical velocity^2 + 57^2

Vertical velocity^2 = 88^2 – 57^2

Vertical velocity = (88^2 – 57^2)^0.5

Tangent of the angle, above the horizontal = (88^2 – 57^2)^0.5 ÷ 57