Question

✏
For an object thrown at 45° to horizontal, the maximum
height (H) and horizontal range (R) are related as
(b) R= 8 H
(d) R= 2 H
(a) R = 16 H
(c) R=4H
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Answer 1

Answer:

Here, angle of projection θ=45oθ=45o

Maximum height H=u2sin2θ2gH=u2sin2⁡θ2g

Horizontal range R=u2sin2θgR=u2sin⁡2θg

For θ=45oθ=45o

H=u2sin245o2g=u24g(∵sin45o=12√)H=u2sin2⁡45o2g=u24g(∵sin⁡45o=12)

R=u2sin90og=u2g(∵sin90o=1)R=u2sin⁡90og=u2g(∵sin⁡90o=1)

∴RH=u2g×4gu2=4∴RH=u2g×4gu2=4

or R=4H

(c) R =4H

Answer 2

Answer:

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