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vaieri [72.5K]
3 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]3 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

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stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

          m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v

Given:

m_{1} = 1090 kg

u_{1} = 11 m/s

u_{2} = 0

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We need to find m_{2}

When substituting the given values in the above equation, we get

(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75

11990=5177.5+4.75 m_{2}

4.75 m_{2}=11990-5177.5

4.75 m_{2}=6812.5

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6 0
2 years ago
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Answer:

\tau=3.96\ ksi

Explanation:

Given that

d= 1.5 in                      ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp                      ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

P=\dfrac{2\pi N\ T}{60}

T=Torque

N=Speed

55927.5=\dfrac{2\times \pi \times 1800\ T}{60}

T=296.85 N.m

The maximum shear stress is given as

\tau=\dfrac{16 T}{\pi d^3}

\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}

\tau=27.35\ MPa

We know that 1 MPa =0.145 ksi

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3 years ago
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Answer:

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We use kinematics to find lips

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