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OverLord2011 [107]
3 years ago
14

To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

ossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 31◦
from the horizontal.
The acceleration of gravity is 9.8 m/s^2.
By how much vertical distance does the ball clear the crossbar?
Answer in units of m.
Physics
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

u_{x}=u\cos\theta

Put the value into the formula

u_{x}=24\cos(31)

u_{x}=20.5\ m/s

We need to calculate the vertical velocity

Using formula of vertical velocity

u_{y}=u\sin\theta

Put the value into the formula

u_{y}=24\sin(31)

u_{y}=12.3\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{u_{x}}

Put the value into the formula

t=\dfrac{44}{20.5}

t=2.1\ sec

We need to calculate the vertical height

Using equation of motion

h=u_{y}t+\dfrac{1}{2}at^2

Put the value into the formula

h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2

h=4.2\ m

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

d=h-3.05

Put the value of h

d=4.2-3.05

d=1.15\ m

Hence, The distance by the ball clear the crossbar is 1.15 m

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Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa
Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

v_r = Speed of river

v_c = Speed of canoe

v_r+v_c=2.8\ m/s

v_r-v_c=-1.9\ m/s

Adding the equations we get

2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s

0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi
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The angular acceleration of a rotating object is given by
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where
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\omega_i is its initial angular speed
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For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
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bagirrra123 [75]

Answer:

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