The vertical height Y and the horizontal distance X along the horizontal plane of projectile .........?

Question

The vertical height Y and the horizontal distance X along the horizontal plane of projectile thrown in air with point of projection as origin are

Y = (8t - 5t2) metre

X = 6t metre

where t is in second.

The velocity with which the body is projected in m/s is

1. 8

2. 6

3. 10

4. cannot be obtained from given data

Answer 1

Vy = Y' = 8 - 10t (m/s)

Vx = X' = 6 (m/s)

Vy(0) = 8 (m/s)

Vx(0) = 6 (m/s)

V0 = √(6^2 + 8^2) = 10 (m/s)

So the answer is 10 (m/s).