Question

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 40 degrees. It reaches the maximum height of 30 m. The acceleration due to gravity is 30 m/s2. Part 1. Find the time t the projectile spends in the air.

Answer 1

Answer:

2.83 s

Explanation:

Let v be the initial launching velocity of the projectile. This can be split into 2 components:

- Horizontal component $v_h = vcos\theta = vcos40^o = 0.766 v$

- Vertical component $v_v = vsin\theta = vsin40^o = 0.643 v$

For the vertical motion, only affected by vertical velocity and gravitational acceleration g = 30 m/s2. As the object reaches maximum height, we can use the following equation of motion to find out the initial velocity of the projectile:

$v^2 - v_0^2 = 2a\Delta s$

where v = 0 m/s is the final velocity of the object when it stops at the max height, $v_0 = v_v = 0.643 v$ is the initial vertical velocity of the object when it start, a = -30 m/s2 is the deceleration of the can, and $\Delta s = 30$ is the maximum distance traveled.

$0 - (0.643v)^2 = 2*(-30)*30$

$(0.643v)^2 = 1800$

$0.643v = \sqrt{1800} = 42.43 = v_v = v_0$

$v = 42.43 / 0.643 = 66 m/s$

Then we can also apply the following equation of motion to solve for the time it takes to reach maximum height of 30m

$\Delta s = v_0t + at^2/2$

$30 = 42.43 t - 30t^2/2$

$15t^2 - 42.43t + 30 = 0$

$t^2 - 2.83t + 2 =0$

As $2.83 = 2*1.415 \approx 2*\sqrt{2}$ we have a case of $(a - b)^2 = a^2 - 2ab + b^2$. The equation of t can be rewritten as

$(t - \sqrt{2})^2 = 0$

$t - \sqrt{2} = 0$

$t = \sqrt{2} \approx 1.415 s$

Since it would take another t seconds to fall down and hit the ground. The total time the projectile would spend on air is 2*1.415 = 2.83 s