Your friend says you can't use the equation x = vo t + 1/2a( t )^2 to find the horizontal displacement of a horizontal projectile with a constant velocity because you don't know the acceleration in the horizontal directon. Is your friend correct? Why or why not? 1. My friend is incorrect. The acceleration is always -9.8 m/s^2
2. My friend is correct. A different equation must be used to find x.
3. My friend is incorrect The horizontal acceleration is the same as horizontal velocity. 4. My friend is incorrect. If the velocity is constant, then the horizontal acceleration is zero.

Explanation:

4. My friend is incorrect. If the velocity is constant, then the horizontal acceleration is zero.

Kinematic equations function when acceleration is constant. Zero qualifies as a constant acceleration.

It's kind of a tricky question because the word "horizontal" presumes there is also a "vertical" which is only valid when in reference to a gravitational field.

A projectile moving initially in a horizontal direction with a vertical gravity acceleration CANNOT maintain a constant velocity. It MAY maintain a constant Horizontal Velocity COMPONENT in the absence of air resistance or other such drag. The Vertical Velocity component will be continuously changing in time due to the gravity acceleration.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west