Question

(problem reference m.3)
a ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. neglect any effects due to air resistance.
what maximum height will the ball reach?
14.7 m
58.8 m
24.5 m
9.80 m
19.6 m

Explanation:

Step1: Identify the kinematic equation

Use the equation \(v^{2}=v_{0}^{2}+2a\Delta y\). At maximum height, \(v = 0\), \(v_{0}=19.6\ m/s\), and \(a=-g=- 9.8\ m/s^{2}\).

Step2: Solve for \(\Delta y\)

Substitute the values into the equation \(0=(19.6)^{2}+2(-9.8)\Delta y\).
First, expand the equation: \(0 = 384.16-19.6\Delta y\).
Then, solve for \(\Delta y\): \(19.6\Delta y=384.16\), so \(\Delta y=\frac{384.16}{19.6}=19.6\ m\).

Answer:

19.6 m