Determining Impact Time of a Ball given Launch Velocity

[rocomath]

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

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

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)

 

[ogb p]

I would like to provide a detailed explanation for the given scenario.

Firstly, it is important to understand that the launch velocity of a ball affects both its horizontal and vertical motion. The horizontal component determines the distance the ball will travel while the vertical component determines the height it will reach.

In the given scenario, the launch velocity of the ball is different in both cases. The first launch velocity is (3m/s) in the horizontal direction and (4m/s) in the vertical direction, while the second launch velocity is (5m/s) in the horizontal direction and (4m/s) in the vertical direction.

Now, when the ball is thrown towards the wall, it will follow a parabolic path due to the influence of gravity. The time taken by the ball to reach the wall will depend on the height at which the ball is thrown, the launch velocity, and the angle at which it is thrown.

In this case, the launch velocities are different, and therefore, the time taken by the ball to reach the wall will also be different. The ball with a higher launch velocity of (5m/s) will cover a greater horizontal distance in the same amount of time compared to the ball with a launch velocity of (3m/s). This means that the ball with a higher launch velocity will reach the wall in a shorter amount of time compared to the ball with a lower launch velocity.

In other words, the time taken by the ball to reach the wall will be less than t_1 when the launch velocity is increased from (3m/s) to (5m/s).

To further convince yourself, you can calculate the magnitude of the launch velocity in both cases. The magnitude of the launch velocity for the first case is √(3^2+4^2)=5m/s, while for the second case it is √(5^2+4^2)=6.4m/s. This clearly shows that the magnitude of the launch velocity is greater in the second case, indicating that the time taken by the ball to reach the wall will also be less.

In conclusion, the time taken by the ball to reach the wall will be less than t_1 when the launch velocity is increased from (3m/s) to (5m/s). Therefore, the correct answer is less than t_1.