Verified answer

Let q = 68 deg, v0 = 35 m/s. Up is the positive y direction.

v0x = v0*cos(q) v0y = v0*sin(q) --> just simple trig here

Motion in x (horizontal) direction follows the formula

x = x0 + v0x*t --> set x0 = 0 since that is a freedom left to you by the problem

vx = v0x --> no acceleration in x direction, speed remains constant

Motion in y is

y = -1/2 g*t^2 + v0y*t + y0 --> g = 9.8 m/s^2 and the "-" indicates it is acting downward (negative y direction)

vy = -1/2 g*t + v0yy

Now at max height, the projectile has no speed in the y direction so the magnitude of the velocity is"

v = sqrt( vx(t)^2 + vy(t)^2) = sqrt( vx(t)^2 + 0) = vx(t) = vx0 = v0*cos(q) = 13.1 m/s

When it hits the ground, it has the same speed as when it was launched (conservation of kinetic and potential energy requires this). Since it left with v = v0 = 35 m/s, the magnitude at impact = 35 m/s. The math to show this is simple.

Find the time to reach max height -- Tmax using vy =0 at t = Tmax

0 = -g*Tmax + v0y --> Tmax = v0y/g

It takes just as long to come back to the ground so the total travel time is T = 2Tmax = 2v0y/g

SInce vx = v0x = constant, that does not change

but vy(T) =-g*T + v0y = -g*2v0y/g + v0y = -v0y

v(T) = sqrt(vx(T)^2 + vt(T)^2) = sqrt(v0x^2 + (-v0y)^2) = sqrt(v0x^2 + (v0y)^2)

= sqrt(v0^2*cos^2(q) + v0^2*sin^2(q)) = v0*sqrt(cos^2(q) +sin^2(q)) = v0 since sin^2(x) + cos^2(x) = 1