Question

Alex climbs to the top of a tall tree while is friend Gary waits on the ground below. Gray throws a ball directly upward to Alex, who allows the ball to go past him before catching the ball on the way down. The ball has an initial speed of 10.0 m/s and is caught 3.50 m above where it was thrown.

a) How long after the ball was thrown, does Alex catch it? Ignore the effects of air resistance.

b) How fast was the ball moving when Alex caught it?

Answer 1

To find the time it takes for Alex to catch the ball, we can use the equation for the vertical motion of an object. After calculating, we find that t ≈ 1.62 seconds. To find the speed of the ball when Alex catches it, we can use the equation v = vi + gt. After calculating, we find that v ≈ -4.16 m/s.

Step-by-step explanation:

To find the time it takes for Alex to catch the ball, we can use the equation for the vertical motion of an object: Δy = vit + (1/2)gt2. In this case, the initial velocity (vi) is 10.0 m/s, the acceleration due to gravity (g) is -9.8 m/s2 (negative because it acts downwards), and the change in height (Δy) is 3.50 m. Plugging these values into the equation, we can solve for t:

3.50 m = (10.0 m/s)t + (1/2)(-9.8 m/s2)t2

This equation is a quadratic equation, so we can solve it using the quadratic formula: t = (-b ± √(b2 - 4ac))/(2a). After substituting in the values and solving, we find that t ≈ 1.62 seconds.

To find the speed of the ball when Alex catches it, we can use the equation: v = vi + gt. Again, the initial velocity is 10.0 m/s and the acceleration due to gravity is -9.8 m/s2. Plugging these values into the equation, we can solve for v:

v = 10.0 m/s + (-9.8 m/s2)(1.62 s)

After calculating, we find that v ≈ -4.16 m/s. Since the ball is caught on the way down, the negative sign indicates that the ball is moving downwards at the time of catching.