Question

Can someone help me?!!!!!

Answer 1

110 m

Step-by-step 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-(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