Question

The position coordinates of a projectile thrown from ground are given by y = 3t – 5t2 (m) and x = 4t (m) (here t is in second, x is horizontal and y is vertical position). horizontal range of the projectile is​

Answer 1

Answer:

If we take x to be the angle of the projection then.

From the question y = 3t - 5t^2 which can be equated with usinxt - gx^2sec^2x/2u^2.

Also, for horizontal projection equating we will get x = ucosxt =4t.

Now, ucosx= 4. tanx = 3/4 as usinx = 4 also

gx^2sec^2x/2u^2 = 5 also sinx=3/5.

So, we get u= 5 m/s.

So, the angle x will be 36 degree.

And value of g will be 32m/s^2.

So, the value of range will be u^2sin2x/g which is 25*0.95/32 which is 0.74 m.