Question

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what is the maximum height reached by the ball? (b) what is the velocity of the ball when it is 96 ft above the ground on its way up? on its way down?

Answer 1

Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16

Answer 2

The kinematics allows finding the answers for throwing the ball upwards are:

      A) The maximum height is 99.94 ft

      B) The time to pass through the height are: for the ascent 2s and for the descent 3s

Kinematics studies the movement of bodies, finding relationships between position, velocity and acceleration.

The reference system is a coordinate system with respect to which to carry out the measurements, in this case we select the positive vertical axis upwards, in the attached we can see a diagram of the launch.

A) They ask for the maximum height at this point the speed must be zero.

     

They indicate the functional relationship of position with respect to time

              y = 80 t - 16 t²

Velocity is defined as the relationship between position and velocity

              v = dy / dt                

             v = 80 - 16 2 t

           

Let's calculate the time

            0 = 80 - 32 t

            t = 80/32

            t = 2.56 s

let's find the position for this time

           y = 80 (2.56) - 16 2.56²

           y = 99.94 ft

B) The time for the height of y = 96 ft

Let's  substitute

           96 = 80 t - 16 t²

Let's solve the quadratic equation

           16 t² - 80 t + 96 = 0

           t² - 5 t + 6 = 0

           t = $\frac{5 \pm \sqrt{5^2 - 4 \ 6} }{2}$  = $\frac{5 \pm 1 }{2}$

           

We have two solutions one for when it is going up and the other for when it is going down

           t₁ = 3s

           t₂ = 2 s

In conclusion, using kinematics we can find the answers for throwing the stone upwards are:

      A) The maximum height is 99.94 ft

      B) The time to pass through the height are: for the ascent 2s and for the descent 3s

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