Physics, 04.07.2020 04:01 chaparro0512

A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 50.0° to the horizontal. Required:
By how much does the ball clear or fall short (vertically) of clearing the crossbar?

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Answer from: huangjianhe135

The difference is height is $\Delta h =6.92 \ m$

Explanation:

From the question we are told that

The distance of ball from the goal is $d = 36.0 \ m$

The height of the crossbar is $h = 3.05 \ m$

The speed of the ball is $v = 21.6 \ m/s$

The angle at which the ball was kicked is $\theta = 50 ^o$

The height attained by the ball is mathematically represented as

$H = v_v * t - \frac{1}{2} gt^2$

Where v_v is the vertical component of velocity which is mathematically represented as

$v_v = v * sin (\theta )$

substituting values

v_v = 21.6 * (sin (50 ))

$v_v = 16.55 \ m/s$

Now the time taken is evaluated as

$t = \frac{d}{v * cos(\theta )}$

substituting value

$t = \frac{36}{21.6 * cos(50 )}$

$t = 2.593 \ s$

$H = 16.55 * 2.593 - \frac{1}{2} * 9.8 * (2.593)^3$

$H = 9.97 \ m$

The difference in height is mathematically evaluated as

$\Delta h = H - h$

substituting value

$\Delta h = 9.97 - 3.05$

$\Delta h =6.92 \ m$

Answer from: Quest

Magma is composed of molten rock and is stored in the earth's crust. lava is magma that reaches the surface of our planet through a volcano vent.

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the cro...