1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula
v² - u² = 2a ∆x
where
u = initial speed
v = final speed
a = acceleration
∆x = displacement
At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so
0² - (20 m/s)² = 2 (-9.8 m/s²) y
Solve for y :
y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m
2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.
3. At any time t ≥ 0, the body's vertical velocity is given by
v = 20 m/s - gt
At the highest point, we have
0 = 20 m/s - (9.8 m/s²) t
and solving for t gives
t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s
4. The body's height y above the ground at any time t ≥ 0 is given by
y = 60 m + (20 m/s) t - 1/2 gt²
Solve for t when y = 0 :
0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²
Using the quadratic formula,
t = (-b + √(b² - 4ac)) / (2a)
(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with
t ≈ 6.09 s
5. At the time found in (4), the body's velocity is
v = 20 m/s - g (6.09 s) ≈ -39.7 m/s
Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.
and the height of the building doesn't matter at all.
joules