Question

the horizontal and vertical x and y of projectile at a given time `t` are given by x=6t metres and y=8t - 5t^2 metres. The range of the projectile in metres is

Answer 1

Answer:

Here's your answer

Answer is 48/5 metres

Answer 2

Answer:

9.6 m.

Explanation:

x = utcos∝ = 6t

y = utsin∝ - 5t² = 8t - 5t²

ucos∝ = 6

usin∝ = 8

u² = 100 , u²sin∝cos∝ = 48 , tan∝ = 4/3

u²sin2∝ = 96

The range of the projectile in metres is = 96/10 = 9.6 m.