A1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. it gets blocked just after reaching a height of 3.0 m.

(a) what is the velocity of the football when it first reaches a height of 3.0 m above the level ground?

(b) what horizontal distance has the ball travelled when it first reaches a height of 3.0 m above ground?

1) Data:

Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m

2) Questions: V at Y = 3m and X at Y = 3 m

3) Calculate components of the initial velocity

Vox = Vo * cos(37°) = 15.97 m/s

Voy = Vo * sin(37°) = 12.04 m/s

4) Formulas

Vx = constant = 15.97 m/s

X = Vx * t

Vy = Voy - g*t

Y = Yo + Voy * t - g (t^2) / 2

5) Calculate t when Y = 3m (first time)

Use g ≈ 9.8 m/s^2

3 = 12.04 * t - 4.9 t^2

=> 4.9 t^2 - 12.04t + 3 = 0

Use the quadratic equation to solve the equation

=> t = 0.28 s and t = 2.18s

First time => t = 0.28 s.

6) Calculate Vy when t = 0.28 s

Vy = 12.04 m/s  - 9.8 * 0.28s = 9.3 m/s

7) Calculate V:

V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s

tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°

V ≈ 18.5 m/s, with angle ≈ 30°

8) Calculate X at t = 0.28s

X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m

X ≈ 4,5 m

Mercury / Mars

Explanation:

For an object launched straight upward, the following SUVAT equation can be used

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity (free fall acceleration) (the negative sign is due to the downward direction of gravity)

h is the maximum height reached

At the maximum height, the velocity is zero, so v = 0. Re-arranging the equation,

So we see that for equal initial velocity (u), the maximum height reaches is inversely proportional to the acceleration of gravity. Therefore, the potato gun will reach the highest altitude in the planets with lowest acceleration of gravity, therefore Mercury and Mars (3.7 and 3.6 m/s^2).