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3 years ago

A ball is launched at 40m/s. find its speed at the half way up point.

1 answer:

6 0

Ek - kinetic energy
v2 - unknown speed
v1 - 40 m/s (initial speed)
Ek=1/2 mv^2
Ek in half way up is 1/2 Ek (with another V)
So, Ek in the beginning is Ek1= 1/2 mv1^2
and in half way Ek2=1/2 mv2^2
Ek1=2*Ek2
1/2 mv1^2 = 2* 1/2 mv2^2
1/2 v1^2 = v2^2
1/2 40^2 = v2^2
800 = v2^2
v2 = sqrt (800) = 28,3 m/s

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<h2>Answer:</h2>

The distance will be 1800 m

<h2>Explanation</h2>

As in question

Time = 15 min

Time = 15 x 60 sec = 900 sec

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Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

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$\huge{\underline{\underline{\red{Answer}}}}$

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