1.
we have given range = 1.40 m
And height = 0.860 m
a) range =( velocity in x direction )× time of flight.....(1)
Time of flight = "\\sqrt{2h\\over g}= \\sqrt{2\u00d70.860\\over 10}" =0.29sec
Using (1)
1.40=speed ×0.29
Speed =4.8 m/s
b)
"V=u+gt"
"V=4.8+10\u00d70.29"
"V=7.7 m\/s"
2. Let angle be "\\theta"
I)max height = "u^2sin^2(\\theta)\\over 2g"
ii)range = "ucos\\theta \u00d7{ 2usin\\theta \\over g}"
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