How to obtain the distance covered by a projectile if an additional acceleration is taken into account?

algebra-precalculus physics kinematics projectile-motion

Solution 1:

When the projectile hits the ground, its velocity is -40m/s, the reverse of the initial velocity,. The time of flight, $t$, is given by the deceleration formula along with the deceleration $-10m/s^2$,

$$(-40)-40=-10t$$

which yields $t=8s$. Then, with the initial horizontal velocity $10m/s$ and the deceleration $-5m/s^2$, the horizontal distance travelled is

$$ d= 10t-\frac 12(5) t^2= 80 - 160 = -80m$$

Thus, the answer is (2). Note that the wind reverses the horizontal travel direction of the projectile and it lends 80m on the opposite side.

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