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

Answer 2

The time spent in the air by the projectile during the entire motion is 2.8 seconds.

The given parameters;

• angle of projection, θ = 40⁰
• maximum height reached the projectile, h = 30 m
• acceleration due to gravity, g = 30 m/s²

The initial velocity of the projectile is calculated as follows;

$v_y_f^2 = v_y_0^2 - 2gh\\\\at \ maximum \ height, \ v_f_y = 0\\\\0 = v_y_0^2 - 2gh\\\\v_y_0^2 = 2gh\\\\(v_0 sin\theta)^2 = 2gh\\\\v_0sin(\theta) = \sqrt{2gh} \\\\v_0 = \frac{\sqrt{2gh} }{sin\theta} \\\\v_0 = \frac{\sqrt{2\times 30 \times 30} }{sin40} \\\\v_0=66 \ m/s$

The time of the projectile to reach the maximum height is calculated as follows;

$h = v_0_yt - \frac{1}{2} gt^2\\\\30 = (66\times sin40)t \ - \ \frac{1}{2}(30)t^2\\\\30 = 42.43t - 15t^2\\\\15t^2 - 42.43 + 30 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 15, \ \ b = -42.43, \ \ c = 30\\\\t = \frac{-b \ \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(42.43) \ \ \ +/- \ \ \sqrt{(42.43)^2 - 4(15\times 30)} }{2(15)} \\\\t = 1.4 \ s$

The total time spent in the air by the projectile is calculated as follows;

T = 2t

T = 2(1.4 s) = 2.8 s.

Thus, the time spent in the air by the projectile during the entire motion is 2.8 seconds.

Learn more here:https://brainly.com/question/8599327