Determining Impact Time of a Ball given Launch Velocity

[rocomath]

You throw a ball with launch velocity of \overrightarrow {v_i}=(3m/s)\widehat{i}+(4m/s)\widehat{j} toward a wall, where it hits at height h_1} in time t_1 after the launch. Suppose that the launch velocity were, instead, \overrightarrow {v_i}=(5m/s)\widehat{i}+(4m/s)\widehat{j}.

Would the time taken by the ball to reach the wall be greater than, less than, or equal to t_1?

I thought that the horizontal component played no role in a projectile motion? Answer is less than, I would have chosen equal to. Should I compute the magnitude in order to convince myself other wise (course it would be greater for the 2nd).

 

[blochwave]

I thought that the horizontal component played no role in a projectile motion?

That's the ONLY component that matters here. Horizontal component is faster, gets there faster. When attempting to figure out how long it takes the ball to travel a horizontal distance, you're only interested in how fast it's going horizontally

Edit: So to make it even clearer, if the problem had the first case being 3 m/s horizontal and 45 sextillion m/s vertically, and the second case is 4 m/s horizontal but 3 nanometers/s vertical, answer is the same(EDIT and by the same I mean the same answer as the original question, which is the second one gets there faster)