A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?


Answer 1


a)the ball will leave the bat at an angle of  61.3°  .

b) the velocity at which it will hit the ground will be v = 27.1 m/s



v = 47.24 m

h = 0.42 m

t = 5.73 s

R = 130 m

a)We know that

R = v cosθ × t

cosθ = (R)/(v t ) = (130)/(47.24* 5.73 ) =0.4803

θ = 61.3°  

the ball will leave the bat at an angle of  61.3°  .

b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s

v = u + at

Vy = 47.24 x sin 61.3 - 9.81 x 5.73

    = -14.8 m/s

v = √(v_x^2 + v_y^2))

v = √(22.7^2 + -14.8^2)

v = 27.1 m/s

the velocity at which it will hit the ground will be v = 27.1 m/s

Related Questions

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A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?
What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.The pressure everywhere increases by the same amount.The pressure everywhere decreases to conserve total pressure.
A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle



The banking angle is 23.98 degrees.


We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :


g is acceleration due to gravity

\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)

So, the banking angle is 23.98 degrees.

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?



a). E_(kA)=1.2635 J

b). V_(B)=4.535(m)/(s)

c). ΔE_(t)=8.4635 J


ΔE=kinetic energy


E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J


E_(kB)=(1)/(2)*m*v_(B) ^(2)

V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)


net work= EkA+EkB

E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J

Omar observes that many buildings in his city were built using limestone. He has read that acid rain can damage limestone. He also knows that limestone reacts with acids and that chemical reactions are affected by temperature. With this in mind, he conducts an investigation to see how the amount of damage to the limestone is affected by the amount of acid on the stone.Which statement describes the correct plan for his procedure?Rhonda watched a video taken by a camera that was lifted into the upper atmosphere by a weather balloon. She saw the balloon pop when it reached a certain height. Afterward, Rhonda wondered what effect the air pressure at high altitudes has on the volume of gas in balloons.

What scientific practice is Rhonda performing


Answer:he conducts the investigation to see the effect acidic water has on limestone


The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.


Given Information:  

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:  

Change in length = ?  

Change in diameter = ?  


Change in length = 0.01257 mm

Change in diameter = -0.003772 mm


Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

Two 3.7 microCoulomb charges are 0.8 m apart. How much energy (in milliJoule) went into assembling these two charges? Enter a number with one digit behind the decimal point.




The potential energy of a system of two charges is given by the expression

(K* Q_1* Q_2)/(R)

Q₁ and Q₂ are two charges and R is distance between the charges.

Given Q₁ = Q₂ = 3.7 X 10⁻⁶ , R = .8 and K = 9 x 10⁹

Putting these values in the equation we have,

[tex](9* 10^9* 3.7*10^(-6) 3.7* 10^(-6))/(.8)[/tex]

Potential energy = 154.01 x 10⁻³ J

This energy have been spent to bring these repelling two charges at this close distance . The energy spent have been stored as potential energy here which has been calculated.

Explain why Planck’s introduction of quantization accounted for the properties of black-body radiation.



The classic model of a black body made predictions of the emission at small wavelengths in open contradiction with what was observed experimentally, this led Planck to develop a heuristic model. This assumption allowed Planck to develop a formula for the entire spectrum of radiation emitted by a black body, which matched the data.