If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related asA. `y = 4h (t/T)(1-t/T)`B. `y = 4h (x/R)(1-x/R)`C. `y = 4h (T/t)(1-T/t)`D. `y = 4h (R/x)(1-R/x)`

If T is the total time of flight, h is the maximum height and R is the

1.	If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related asA. `y = 4h (t/T)(1-t/T)`B. `y = 4h (x/R)(1-x/R)`C. `y = 4h (T/t)(1-T/t)`D. `y = 4h (R/x)(1-R/x)`
Answer» Correct Answer - A::B a,b. `T = (2usintheta)/g , h = (u^2sin^2theta)/(2g), R = (u^2sin2theta)/g` `y = usinthetat - 1/2 "gt"^2` `=4ht[(usintheta)/(4h) - ("gt")/(8h)]` `= 4ht[ (usintheta)/(4(u^2sin^2theta//2g)) - ("gt")/(8(u^2sin^2theta//2g))]` `= 4ht[1/(2usintheta//g) - t/((2usintheta//g)^2)]` `4ht[1/T - t/T^2] = 4h (1/T)(1-t/T)` Now `x = ucos theta t, R = u cos theta T` `rArr t/T = x/R, so y = 4h(x/R)(1-x/R)`.

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