Question

You toss a ball straight up with an initial speed of 22 m/s. How high does it go?

a. 20 m
b. 45 m
c. 50 m
d. 110 m

Answer 1

When a ball is tossed straight up with an initial speed of 22 m/s, the maximum height it can reach is calculated using the formula h = (v^2 - u^2) / (2g), resulting in roughly 24.7 meters. The closest answer option given is 20 meters (a).

Step-by-step explanation:

To calculate how high a ball goes when it's tossed straight up with an initial velocity, we can use the physics equation for displacement with constant acceleration. The formula is:

• h = (v^2 - u^2) / (2g)

where h is the maximum height, v is the final velocity (0 m/s at the highest point), u is the initial velocity (22 m/s in this case), and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging the values into the formula, we get: h = (0^2 - 22^2) / (2 * -9.8), which simplifies to h = 484 / 19.6 = 24.7 m.

So the maximum height the ball will reach is approximately 24.7 meters. The closest answer provided is option a, 20 m.