Can some one please explain how to do this? 1.253 x 7.4 =

У-3

alisha [4.7K]

Answer:

the answer will be verified by an expert. Until then, talk to a tutor.

5 0

2 years ago

3. A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.

Vikki [24]

Let the rocket is launched at an angle of \theta with respect to the positive direction of the x-axis with an initial velocity u=15m/s.

Let the initial position of the rocked is at the origin of the cartesian coordinate system where the illustrative path of the rochet has been shown in the figure.

As per assumed sine convention, the physical quantities like displacement, velocity, acceleration, have been taken positively in the positive direction of the x and y-axis.

Let the point P(x,y) be the position of the rocket at any time instant t as shown.

Gravitational force is acting downward, so it will not change the horizontal component of the initial velocity, i.e. $U_x=U cos\theta$ is constant.

So, after time, t, the horizontal component of the position of the rocket is

$x= U \cos\theta t \;\cdots(i)$

The vertical component of the velocity will vary as per the equation of laws of motion,

$s=ut+\frac12at^2\;\cdots(ii)$ , where,s, u and a are the displacement, initial velocity, and acceleration of the object in the same direction.

(a) At instant position P(x,y):

The vertical component of the initial velocity is, $U_y=U sin\theta$ .

Vertical displacement =y

So, s=y

Acceleration due to gravity is $g=9.81 m/s^2$ in the downward direction.

So, $a=-g=-9.81 m/s^2$ (as per sigh convention)

Now, from equation (ii),

$y=U sin\theta t +\frac 12 (-9.81)t^2$

$\Rightarrow y=U \sin\theta \times \frac {x}{U \cos\theta} +\frac 12 (-g)\times \left(\frac {x}{U \cos\theta} \right)^2$

$\Rightarrow y=U^2 \tan\theta-\frac 1 2g U^2 \sec^2 \theta\;\cdots(iii)$

This is the required, quadratic equation, where $U=15 m/s$ and $g=9.81 m/s^2$ .

(b) At the highest point the vertical velocity,v, of the rocket becomes zero.

From the law of motion, $v=u+at$

$\Rightarroe 0=U\sin\theta-gt$

$\Rightarroe t=\frac{U\sin\theta}{g}\cdots(iv)$

The rocket will reach the maximum height at $t= 1.53 \sin\theta s$

So, from equations (ii) and (iv), the maximum height, $y_m$ is

$y_m=U\sin\theta\times \frac{U\sin\theta}{g}-\frac 12 g \left(\frac{U\sin\theta}{g}\right)^2$

$\Rightarrow y_m=23 \sin\theta -11.5 \sin^2\theta$

In the case of vertical launch, $\theta=90^{\circ}$ , and

$\Rightarrow y_m=11.5 m$ and $t=1.53 s$ .

Height from the ground= 11.5+3=14.5 m.

(c) Height of rocket after t=4 s:

$y=15 \sin\theta \times 4- \frac 12 (9.81)\times 4^2$

$\Rightarrow y=15 \sin\theta-78.48$

$\Rightarrow -63.48 m >y> 78.48$

This is the mathematical position of the graph shown which is below ground but there is the ground at $y=-3m$ , so the rocket will be at the ground at t=4 s.

(d) The position of the ground is, y=-3m.

$-3=U\sin\theta t-\frac 1 2 g t^2$

$\Rightarrow 4.9 t^2-15 \sin\theta t-3=0$

Solving this for a vertical launch.

$t=3.25 s$ and $t=-0.19 s$ (neglecting the negative time)

So, the time to reach the ground is 3.25 s.

(e) Height from the ground is 13m, so, y=13-3=10 m

$10=U\sin\theta t-\frac 1 2 g t^2$

Assume vertical launch,

$4.9 t^2-15 \sin\theta t+10=0$ [using equation (ii)]

$\Rightarrow t=2.08 s$ and $t=0.98 s$

There are two times, one is when the rocket going upward and the other is when coming downward.

4 0

3 years ago

HELP

Reptile [31]

Answer:

The time it takes for the cannonball to hit the ground is 6.49 seconds.

Step-by-step explanation:

You know that the height of the cannonball above the ground, h, in meters, in time, t, in seconds, is found by the function h(t) = -4.9t² + 30.5t + 8.4. You want to calculate the time it takes for the cannonball to hit the ground, that is, the time it takes for the bullet to reach zero height. So, being h(t) = 0 you have:

0 = -4.9t² + 30.5t + 8.4

To solve a quadratic function 0=a*x² + b*x +c, the expression is applied:

$\frac{-b+-\sqrt{b^{2} -4*a*c} }{2*a}$