As velocity__kinetic energy _ and potential energy ___

Physics

1 answer:

nataly862011 [7]1 year ago

6 0

Answer:

increases increases decreases

Explanation:

For an object falling from a height

You might be interested in

Please help. I don’t understand this

skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

v (ave) = ∆x / ∆t

and under constant acceleration,

v (ave) = (v (final) + v (initial)) / 2

According to the plot, with ∆t = 4.0 s, we have v (initial) = 0 and v (final) = 10.0 m/s, so

∆x / (4.0 s) = (10.0 m/s) / 2

∆x = ((4.0 s) • (10.0 m/s)) / 2

∆x = 20.00 m

5 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

2-D Projectile Motion

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$