All categories

3 years ago

A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

Physics

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t :

Equations of the uniformly accelerated rectilinear motion of upward (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the initial vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 = (v₀y)²

$v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }$

v₀y = 14.3 m/s

Calculation of the maximum height the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball (R)

We replace data in the equation (1)

x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

$v= \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v= \sqrt{(7.7)^{2}+ (-14.3)^{2} }$

v= 16,2 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })$

$\alpha = tan^{-1} (\frac{-14.3 }{7.7 })$

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

You might be interested in

6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

gavmur [86]

You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.

5 0

3 years ago

Read 2 more answers

Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

vlabodo [156]

GAVNSTVVHUYCZ swffggff

4 0

3 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago

Upthrust given by Water to the body

erica [24]

Answer:

According to Archimedes principle the upthrust on the body is equal to the weight of the water displaced by the body. ... Here, the mass would be the net difference in the weight of the object.

7 0

3 years ago

seropon [69]

Answer: Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic energy increases by a factor of four. Kinetic energy is proportional to the square

of the velocity. If the velocity of an object

doubles, the kinetic energy increases by a

factor of four.

• Kinetic energy is proportional to the mass. If

a bowling ball and a ping pong ball have the

same velocity, the bowling ball has much

larger kinetic energy.

• Kinetic energy is always positive.

• unit : Joule (J) = kg m

2 Example:

If we drop a 3-kg ball from a height of h = 10 m,

the velocity when the ball hits the ground is

given by: v 2 = v0 2 +2a(y− y0 )= 0−2g(0−h)v= 2gh= 2(9.8 m/s 2 )(10 m)=14 m /s Initial: k = 1 2 mv 2 = 0 Final: k = 1 2 mv 2= 1 2 (3 kg)(14 m/s) 2= 294 J So as the ball falls, its kinetic energy increases. It is the gravitational force that accelerates the ball, causing the speed to increase. The increase in speed also increases the kinetic energy. The process of a force changing the kinetic energy of an object is called work. Work: Work is the energy transferred to or from an object by mean of a force acting on the object.• energy transferred to an object is positive work, e.g. gravity performs positive work on a

falling ball by transferring energy to the ball, causing the ball to speed up.• energy transferred from an object is negative work, e.g. gravity performs negative work on a ball tossed up by transferring energy from the ball, causing the ball to slow down.• both kinetic energy and work are scalars.• unit: J Work Energy Theorem: The work done is equal to the change in the kinetic energy: ∆K = K f − K i = W In the above example with the ball falling from a height of h = 10 m, the work done by gravity: W = ∆k = k f −k i = 294 J− 0J = 294 J. If a ball rises to a height of h =10 m, the work done by gravity: W = ∆k = k f −k i = 0J−294 J = −294 J. Work Done by a Force: Consider a box being dragged a distance d across a frictionless floor:

d F y x θ v 2 = v0 2 + 2ax (x − x0 ) v 2 = v0 2 +2ax d 1 2 mv 2 = 1 2 mv0 2 +max d 1 2 mv 2 − 1 2 mv0 2 = max d k f −k i = (Fcosθ)d ∴W = (Fcosθ)d• θ is the angle between the force vector and the direction of motion.• If the force is perpendicular to the direction of motion, then the work done: W =(Fcosθ)d = Fdcos90°= Fd×0= 0.• The work energy theorem and the relationship between work and force are valid only if the force does not cause any other form of energy to change, e. g. we can not apply the theorem when friction is

involved because it causes a change in the thermal energy (temperature). Work Done by Multiple Forces: The total work done by many forces acting on an object:Wtot = F1 cosθ 1 d+F2 cosθ 2 d+ F3 cosθ 3 d+L where the angles are the angle between each force and the direction of motion. The total work is just the sum of individual work from each force:Wtot =W1 +W2 +W3 +L The work energy theorem relates the changes in the kinetic energy to the total work performed on the object: ∆K =Wtot Example: A 3-kg box initially at rest slides 3 m down a frictionless 30° incline. What is the work done on the object? What is the kinetic energy and speed at the bottom?

x y N φ φ mg• The work done is performed by the force in the x direction since there is no motion in the y direction: W = F x d =(mgsinφ)d =(3 kg)(9.8 m/s 2 )(sin30°)(3 m) = 44 J Alternatively, W =(Fcosθ)d = Fcos(90°−φ)d = FsinφdH The first method of using the component of the force in the direction of motion for the calculation is easier.

8 0

3 years ago