A ball thrown up vertically returns to the thrower after 6 s. ... - Science

Given:

• Total time (T) = 6 \text{ s}

• Time to reach maximum height (t) = 6/2 = 3 \text{ s}

• Final velocity at max height (v) = 0 \text{ m/s}

• Acceleration due to gravity (g) = -9.8 \text{ m/s}^2 (upward motion)

(a) The velocity with which it was thrown up (u):
Using the equation: v = u + gt
0 = u + (-9.8 \times 3)
0 = u - 29.4
u = 29.4 \text{ m/s}

(b) The maximum height it reaches (h):
Using the equation: s = ut + \frac{1}{2}gt^2 (taking t = 3 \text{ s})
h = 29.4(3) + \frac{1}{2}(-9.8)(3)^2
h = 88.2 - 4.9(9)
h = 88.2 - 44.1
h = 44.1 \text{ m}

(c) Its position after 4 s:

Using the equation: 

s = ut + \frac{1}{2}gt^2 (taking t = 4 \text{ s})
s = 29.4(4) + \frac{1}{2}(-9.8)(4)^2
s = 117.6 - 4.9(16)
s = 117.6 - 78.4
s = 39.2 \text{ m}

Answer: The ball is at a height of 39.2 m from the ground after 4 seconds.