Question

The height y and horizontal distance x covered by a projectile in a time t seconds are given by the
equations y = 8+ - 512 and x = 6t. If x. and y are measured in metres, the velocity of projection is​

Answer 1

Answer:

10m/s^2

Explanation:

x=6t

on differentiating with respect to t

vx=6

y=8t−5t^2

vy  =8−5t

at t=0 vy  =8

v^2 =6^ 2  +8^2  =10^ 2

v=10m/s^2

Answer 2

Explanation:

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