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a body is projected in a vertical direction.... the altitude y is given by y= 10tsquare-9t + 5 calculate the initial velocity..
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Answered by
4
Answer:
Y = 8t - 5t²
differentiate wrt time
dy/dt = 8 -10t
we know, change in position per unit time is called , velcity .
e.g dy/dt = Vy
Vy = 8 -10t
now at t = 0 ( initially )
Vy =8 -10× 0 = 8
x = 6t
differentiate wrt time
dx/dt = 6
similarly , dx/dt = Vx =6
now,
V = Vx i + Vy j
= 8 i + 6 j
magnitude of velocity =| V | =√(8² + 6²) =10 m/s
put your values and solve in same way...
#gundi..✌✌
Answered by
2
1 erg/ second= 1.0× 10 to the power 7 watts
2 \times {10}^{9} \times 1.0 \times {10}^{7}2×10
9
×1.0×10
7
200 watts ans
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