Question

two particals projected with same speed in different directions and with same vertical plane . both partical have same range, the time taken are 10sec and 25sec find their horizontal range​

Answer 1

Answer:

We know that the range R=u

2

Sin2θ/g is same for angles θ and 90−θ provided projection speed is same.

Suppose projection speed is u then

time of flight from T=2uSinθ/g is obtained as T

1

=2uSinθ/g and T

2

=2uSin(90−θ)/g=2uCosθ/g

On dividing we get T

1

/T

2

=10/25=0.4=tanθ so θ=tan

−1

0.4=21.8

0

putting this value in expression for T

1

we get u=134.77m/s

So range will be R=u

2

Sin2θ/g=134.77

2

×0.6896/10=1252meter

Explanation:

HOPE IT HELPED

MARK AS BRAINLIEST

Answer 2

Answer:

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