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2 years ago

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A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. If the pass w

as thrown with a 14.2 degree angle, with an initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used, how long did this pass take in seconds?

1 answer:

Sauron [17]2 years ago

4 0

<span>This projectile has been thrown in an oblique movement, so it’s moving in x and in y at the same time. Vy varies due to gravity, and Vx is a constant as there is no gravity in the horizontal direction. In order to calculate Vx, you have to use the angle: Vx=V*cosa=12m/s*cos(14.2)= 11.6m/s So now you should use Vx the same way it is used in a uniform linear movement. d=v*t dx=vx*t t=dx/vx=7m/(11.6m/s)=0.603s</span>

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A student bikes to school by traveling first dN = 1.00 miles north, then dW = 0.600 miles west, and finally dS = 0.200 miles sou

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Refer to the diagram shown below.

Define unit vectors along the x and y axes as respectively $\hat{i} \, and \, \hat{j}.$

Then the three successive displacements, written in component form, are respectively
$\vec{dN} = 1.0 \, \hat{j} \\
\vec{dW} = -0.6 \, \hat{i} \\ \vec{dS} = -0.2 \, \hat{j}$

The total displacement for the first leg of the trip is
$\vec{d} = \vec{dN} + \vec{dW} + \vec{dS} \\ \vec{d}= 1.0\hat{j}-0.6\hat{i}-0.2\hat{j} \\ \vec{d}=-0.6\hat{i}+0.8\hat{j}$

Answer:
$\vec{d} = -0.6\hat{i}+0.8\hat{j}$ or (-0.6, 0.8)

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2 years ago

A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(

maks197457 [2]

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

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2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago