A ball is thrown upward from a height of 2.3 meters at an initial speed of 58 m/s^2 . Acceleration resulting from gravity is -9.8m/s^2.

Question

Part 1:Neglecting air resistance, solve for the velocity, v(t), and the height, h(t), of the ball t seconds after it is thrown and before it hits the ground.

Part 2:The distance traveled by the ball between 2 seconds and 9 seconds is___m?

Answer 1

Using kinematics equations

h(t) = h₀ + v₀t + (1/2)at²

dh(t)/dt = v₀ + at

Plug in your given values

h(t) = 2.3 + 58t - (1/2)(9.81)t²

v(t) = 58 - 9.81t

Part (b)

First find time of max height, this is when v(t) = 0

v(t) = 58 - 9.81t = 0, solve for t

t = 5.912 seconds

So from t = 2 seconds to t = 5.912 seconds, it is increasing in height

Now find the time when the ball hits the ground, this is when h(t) = 0

h(t) = 2.3 + 58t - 4.905t² = 0, solve for t using quadratic formula

t = -0.040 and 11.864 seconds, ignore the negative value.

So from t = 5.912 seconds to t = 11.864 seconds, it is decreasing in height.

Next, calculate the max height, this is when t = 5.912 seconds

h(5.912) = 2.3 + 58(5.912) - 4.905(5.912)²

^{h = 173.758 m}

^{So from t = 2 seconds to t = 5.912 seconds, the ball travels a distance of 173.758 - 2.3 = 171.458 m}

Next calculate height at t = 9 seconds

h(9) = 2.3 + 58(9) - 4.905(9)² = 126.995 m

So from t = 5.912 seconds to t = 9 seconds, the ball travels a distance of 173.758 - 126.995 = 46.763 m

Add up both distances and this is the total distanced traveled

171.458 + 46.763 = 218.221 m

Hope this helps