1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vaieri [72.5K]
4 years ago
5

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

rows the softball with a velocity of
23.5
m/s
at an angle of
33.5
∘
above the horizontal. When the softball leaves her hand, it is
16.5
m
above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Physics
1 answer:
Andre45 [30]4 years ago
4 0

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

  • A uniform motion along the horizontal (x) direction
  • A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

where

u = 23.5 m/s is the initial velocity

\theta=33.5^{\circ} is the angle of projection

Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s

Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

You might be interested in
You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time
irinina [24]
It is -59 i think or just ask someone esle for help
8 0
3 years ago
A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.
leonid [27]
The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m
7 0
3 years ago
An 100W light bulb is on 6 hours work out the energy (kWh) used and the cost of it (1kWh = £0.1359) with working out please
Ainat [17]
1,000 W  =  1 kW
100 W  =  0.1 kW

           (0.1 kW) x (6 h) = 0.6 kWh    <=== energy

           (0.6 kWh) x (£0.1359/kWh)  =  £0.0815     <=== cost of it     
8 0
3 years ago
A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Ksju [112]

Answer:

M_1 = 317.7 kg

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

m = \rho V

m = 0.179(\frac{4}{3}\pi R^3)

m = 0.179(\frac{4}{3}\pi 6.55^3)

m = 210.7 kg

now total mass of balloon + helium inside balloon is given as

M = 210.7 + 990

M = 1200.7 kg

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

(M + M_1)g = \rho_{air} V g

(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)

1200.7 + M_1 = 1518.4

M_1 = 317.7 kg

3 0
4 years ago
If a train travels 500 kilometers from Stockholm to
o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0
1 year ago
Other questions:
  • The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio
    6·1 answer
  • Please help me asapp!!!!!!!!!!!!!!!!
    10·2 answers
  • The image shows a model of the gravitational field around Earth. Which best describes the model?
    12·2 answers
  • What are two main factors that affect how quickly a coastline erodes?
    14·1 answer
  • Will is a scientist. He’s designing a spacecraft that would allow people to land on Mars. Will’s mass on Earth is 75 kilograms.
    6·2 answers
  • What are four things that are cycled in the environment?
    7·2 answers
  • What orientation of the dipole has the greatest electric potential energy? What orientation of the dipole has the greatest elect
    7·1 answer
  • The _____ is the process scientists use to conduct research, which includes a continuing cycle of exploration, critical thinking
    13·2 answers
  • What property of objects is best measured by their capacitance?
    11·1 answer
  • The place below earth's surface where the earthquake begins.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!