Question

Can someone help me?!!!!!

Answer 1

Answer:

110 m

Explanation:

First of all, let's find the initial horizontal and vertical velocity of the projectile:

$v_{x0}=v cos 30^{\circ}=(25 m/s)(cos 30^{\circ})=21.7 m/s$

$v_{y0}=v sin 30^{\circ}=(25 m/s)(sin 30^{\circ})=12.5 m/s$

Now in order to find the time it takes for the projectile to reach the ground, we use the equation for the vertical position:

$y(t)=h+v_{0y}t-\frac{1}{2}gt^2$

where

h = 65 m is the initial height

t is the time

g = 9.8 m/s^2 is the acceleration due to gravity

The time t at which the projectile reaches the ground is the time t at which y(t)=0, so we have:

$0=65+12.5 t - 4.9t^2$

which has 2 solutions:

t = -2.58 s

t = 5.13 s

We discard the 1st solution since its negative: so the projectile reaches the ground after t=5.13 s.

Now we know that the projectile travels horizontally with constant speed

$v_x = 21.7 m/s$

So, the horizontal distance covered (x) is

$x=v_x t = (21.7 m/s)(5.13 s)=111.3 m$

So the closest option is

110 m