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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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0.54 A

Explanation:

Parameters given:

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Area of coil, A = 40 cm² = 0.004 m²

Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T

Time interval, Δt = 2 secs

Resistance of the coil, R = 0.2 ohms

To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:

|V| = |(-N * ΔB * A) /Δt)

|V| = | (-15 * 3.6 * 0.004) / 2 |

|V| = 0.108 V

According to Ohm's law:

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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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