The height (H) of the ball thrown into the air with an initial velocity of 9.8 m/s from a height of 2 m above the ground is given by the..?

a.
Method 1 :

H(t) = -4.9t² + 9.8t + 2
H(t) = -4.9(t² - 2t) + 2
H(t) = -4.9(t² - 2t + 1) + 4.9 + 2
H(t) = 6.9 - 4.9(t - 1)²

For all real t, -4.9(t - 1)² ≤ 0
Then, H(t) = 6.9 - 4.9(t - 1)² ≤ 6.9

The maximum height reached by the object = 6.9 m


Method 2 :

H(t) = -4.9t² + 9.8t + 2
H'(t) = -9.8t + 9.8 = -9.8(t + 1)
H"(t) = -9.8

When t = 1 :
H(1) = -4.9(1)² + 9.8(1) + 2 = 6.9
H'(t) = 0
H"(t) = -9.8 < 0

The maximum height reached by the object = 6.9 m


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b.
6.9 - 4.9(t - 1)² = 6.9
(t - 1)² = 0
t = 1

Time taken for the ball to reach the maximum height = 1 s


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c.
H(t) = 4
-4.9t² + 9.8t + 2 = 4
4.9t² - 9.8t + 2 = 0
t = [9.8 ± √(9.8² - 4*4.9*2)] / (2*4.9)
t = 1.77 or t = 0.23

Time taken = 0.23 s or 1.77 s

applying perhaps a more Physics oriented approach

H = maximum height reached (from horizontal level of point of projection) = [initial vertical velocity]²/[2g]

if g ~= 9.8 m/s² = acceleration due to gravity

=> H = 9.8² / [2x9.8] = 4.9 m

from the ground level maximum height = 4.9 + 2 = 6.9 m (Ans)

time taken to reach maximum height from projection instant = [initial vertical velocity]/[g] = u/g = 9.8/9.8 = 1 s

c) if H(t) = -4.9t² + 9.8t + 2 = 4 m

=> H(t) = -4.9t² + 9.8t - 2 = 0 or => t² - 2t + 0.4082 = 0

solving => t = 1.769 s or 0.231 s

explanation is that the ball first reaches a height of 4 m above the ground on its ascent to the high point(6.9 m). its velocity is then 0(momentarily at rest) which in turn changes direction(sign) to return groundwards. on its descent it reaches a 4 m height totally after 1.769 s

it first reaches 4m height(in ascent) after 0.231 s


hope this helps