Application of Antiderivatives in PVA (position, velocity, acceleration)

Question

A child throws a ball straight up. If the ball is to reach a height of 9.8m, what is the minimum initial velocity that must be imparted to the ball? Assume the initial height of the ball is 1m. Assume acceleration due to gravity is -9.8 m/s^2. Solve this question using antiderivatives. (My teacher asked this question as test prep. None of our classmates understand it, can someone help?) Thanks!

Answer 1

Start with three basic things:

1) a = -9.8 (units are m/s²)

2) a = dv/dt

3) v = dy/dt (assuming vertical motion is "y")

Since a = -9.8 then:

dv/dt = -9.8

dv = -9.8 dt

∫dv = ∫-9.8 dt

v = -9.8t + C

Let v₀ equal the velocity at time t = 0 then C = v₀ so:

v(t) = -9.8t + v₀

Since v = dy/dt then:

dy/dt = -9.8t + v₀

dy = -9.8t + v₀ dt

∫dy = ∫-9.8t + v₀ dt

y = (1/2)(-9.8)t² + v₀t + C

Let y0 equal the position at time t = 0 then C = y₀ so:

y(t) = -4.9t² + v₀t + y₀

We want y(t) = 9.8 and we are given y₀ = 1 so plugging those into this equation we get:

9.8 = -4.9t² + v₀t + 1

Using the velocity equation:

v(t) = -9.8t + v₀ consider that at the maximum height, the velocity will be instantaneously be zero so:

0 = -9.8t + v₀

t = v₀/9.8 so plug "v₀/9.8" into the "9.8 = -4.9t² + v₀t + 1" in place of "t":

9.8 = -4.9(v₀/9.8)² + v₀(v₀/9.8) + 1

8.8 = -4.9(v₀/9.8)² + v₀(v₀/9.8) now multiply both sides by 9.8 to get:

86.24 = -0.5v₀² + v₀²

86.24 = 0.5v₀²

172.48 = v₀²

v₀ = √172.48 = 13.13 m/s

Answer 2

height function is

h(t) =-4.9t^2 +vot +1 where t= time measured in seconds

it reaches max height when h'=0 = v(t)= -9.8t +vo = 0 = velocity at time t

vo = initial velocity

= 9.8t

find minimum v such that h = at least 9.8 meters

vo^2 = 2gh = 2(9.8)(9.8-1)

vo=sqr(2(9.8)(8.8)) = about 13.13 m/sec

t=vo/9.8= about 13.13/9.8 = 1.34 seconds

check the answer, plug in those vo and t values into the original height function

h(1.34)=-4.9(1.34)^2 + 13.13(1.34)+1=about 9.8