Physics, 09.10.2019 07:30 goblebrandon

One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
while your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
assume both snowballs are thrown with
the same initial speed 20.5 m/s. the first
snowball is thrown at an angle of 68◦
above
the horizontal. at what angle should you
throw the second snowball to make it hit the
same point as the first? the acceleration of
gravity is 9.8 m/s

the answer is: 22

part 2:
how many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
answer in units of s.
find out part 2

Answers: 1

Show answers

Answers

Answer from: sandrahg3

Part One: $22^{\circ}$ ;

Part Two: Toss the snowball approximately 2.31 seconds after tossing the first snowball.

Assumption: there's no air resistance on the snowballs. Both snowballs stay intact during the flight.

Explanation:

Part OneInitial vertical velocity of the first snowball: $20.5\sin 68^{\circ} \approx \rm 19.0\; m\cdot s^{-1}$ ;Horizontal velocity of the first snowball: $20.5\cos68^{\circ} \approx \rm 7.68\;m \cdot s^{-1}$ .

How long will the first snowball stay in the air?

The ball reaches its vertex (highest point in the trajectory, where vertical velocity is zero) at $\displaystyle \frac{ 19.0}{9.81}\; \rm s$ .It will take the ball another $\displaystyle \frac{ 19.0}{9.81}\; \rm s$ to land. In other words, the first snowball stays in the air for $\displaystyle 2\times \frac{ 19.0}{9.81}\; \approx 3.97\rm s$ .

What will be the range of the first snowball? In other words, how far will the first ball travel horizontally before it lands?

$3.97 \times 7.68 \approx 30.5\; \rm m$ .

That should also be the range of the second snowball.

Let the angle of elevation of the second snowball at takeoff be $\theta$ .

Initial vertical velocity of the second snowball: $20.5\sin \theta$ ;Horizontal velocity of the second snowball: $20.5\cos\theta$ .

The second snowball will spend

$\displaystyle \frac{30.5}{20.5\cos\theta}$ seconds

in the air before covering that range of approximately $30.5\; \rm m$ .

It would take half that much time for the ball to reach its vertex where its vertical velocity equals zero.

$\displaystyle \frac{20.5\sin \theta}{9.8} = \frac{1}{2}\cdot\frac{30.5}{20.5\cos\theta}$ .

Solve this equation for the angle $\theta$ .

Multiply both sides by $2\times 9.8\times 20.5\cos\theta$ :

$20.5^{2}(2\sin\theta \cos\theta)= 30.5\times 9.8$ .

Apply the double-angle identity for the sine of an angle:

$2\sin\theta \cos\theta = \sin 2\theta$ :

$20.5^{2} \sin2\theta= 30.5\times 9.8$ .

The result shall be between $22^\circ$ and $23^\circ$ . For a more precise value, keep more significant figures during the calculation.

Part Two

Recall that it is concluded in part one that the second snowball will spend

$\displaystyle \frac{30.5}{20.5\cos\theta}$ seconds

in the air before covering that range of approximately $30.5\; \rm m$ (or more precisely, $29.7888\rm \; m$ .)

$\theta = 22^\circ$ . The second snowball will spend $\displaystyle \frac{29.7888}{20.5\cos22^{\circ}}\approx 1.5672\; \rm s$ in the air.

The first snowball will spend around $3.8790\rm \; s$ . The difference between the two time is the number of seconds that the person throwing the snowball need to wait:

$3.8790 - 1.5672 \approx 2.31\rm \; s$ .

Answer from: Quest

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample, $v = 1.1 cm^{3}$

so, density of rock sample will be

$density = \frac{mass}{volume}$

$\rho =\frac{2.7}{1.1}g/cm^{3}\\\\\rho =2.45 g/cm^{3}$

since, the density of rock sample is $2.45 g/cm^{3}$ . hence, the rock sample is a sedimentary rock, whose densities varies between $2g/cm^{3} - 2.6 g/cm^{3}$ .