Question

An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vo and second time he increases the initial sped to 3vo . How would you compare the maximum height in the second trial to that in the first trial?

Answer 1

The maximum height reached by an arrow when shot vertically with initial speed 3vo is 9 times the height it reaches with an initial speed vo.

Step-by-step explanation:

The question involves comparing the maximum height reached by an arrow when shot straight up with two different initial speeds: vo and 3vo. According to the kinematic equations for projectile motion, the maximum height H that an object reaches when projected vertically is given by H = v2 / (2g), where v is the initial vertical velocity and g is the acceleration due to gravity.

For the first shot, the maximum height is H1 = vo2 / (2g). For the second shot with initial speed 3vo, the maximum height is H2 = (3vo)2 / (2g) = 9vo2 / (2g). This simplifies to H2 = 9 * H1. Thus, the maximum height in the second trial is 9 times the maximum height in the first trial.

The maximum height reached by a projectile depends only on its vertical component of velocity. When an archer shoots an arrow straight up, the initial speed determines the maximum height it will reach. In the first trial, the initial speed is vo, so the maximum height reached will be H1.

In the second trial, the initial speed is increased to 3vo, so the maximum height reached will be H2. Since the maximum height depends on the vertical component of velocity, H2 will be greater than H1. Therefore, the maximum height reached in the second trial is higher than that in the first trial.