Qui:

A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

4 years ago

Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i

LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

$v^{2} = u^{2} + 2as$

a = $v^{2} - 2gh$

or, h = $\frac{v^{2}}{2g}$

= $\frac{(26)^{2}}{2 \times 9.8}$

= $\frac{676}{19.6}$

= 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

v = u + at

or, v - gt = 0

t = $\frac{v}{g}$

= $\frac{26}{10}$

= 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c) Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d) Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

3 0

3 years ago

Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the

ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N

4 0

3 years ago

An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately 12.77 seconds.

c) The speed of the package as it lands is approximately 145.51 m/s.

d) The path of the package based on a stationary frame of reference is parabolic

e) The path of the package as seen from the plane is directly vertical downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

$\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0

2 years ago

How many miles is in 7 blocks?

borishaifa [10]

Answer:

Different street blocks are different lengths, so it won't be possible to answer this.

8 0

3 years ago