Question

A ball is thrown upward from a height of 15 m with a velocity of 20 m/sec. Acceleration due to gravity is 9.8 m/s2. A. Find the relation between height h and time t after the ball is released. B. How high is the ball after 3 seconds?C. When will the ball hit the ground?2. Repeat problem 1, only answer the questions as if the ball were on the moon. Acceleration due to gravity on the moon is 1.6 m/s2. 3. A ball is kicked upward from a height of 1 m with a velocity of 25 m/sec. Acceleration due to gravity is 9.8 m/s2a. Find the relation between height h and time t after the ball is released.B. How high is the ball after 2 seconds?C. When will the ball hit the ground?D. What is the maximum height of the ball?

Answer 1

A. h = h₀ + u·t - 1/2·g·t²

B. 30.9 m

C. 4.73 seconds

2. A. h = h₀ + u·t - 1/2·a·t²

B. 67.8 m

C. Approximately 25.73 seconds

3. A. h = h₀ + u·t - 1/2·g·t²

B. 31.38 m

C. Approximately 5.142 seconds

D. Approximately 32.9 m

Explanation:

The given parameters are;

The initial height of the ball, h₀ = 15 m

The upward velocity with which the ball is thrown, u = 20 m/sec.

The acceleration due to gravity, g = 9.8 m/s²

A. The relation between the height, h, and the time, t, after the ball is released is given as follows;

h = h₀ + u·t - 1/2·g·t²

B. The height of the ball after 3 seconds is given by substitution as follows;

At t = 3 seconds, h = 15 + 20 × 3 - 1/2 × 9.8 × 3² = 30.9

The height of the ball, h, after 3 seconds is h = 30.9 m

C. The time the ball takes to hit the ground = 2 × The time it takes to maximum height + The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height

The time it takes to maximum height,
`t_(max)`, is given as follows;

v = u - g·
`t_(max)`

Where;

v = The final velocity = 0 at maximum height

Therefore, we have;

0 = 20 - 9.8 ×
`t_(max)`

∴
`t_(max)` = 20/9.8 ≈ 2.0408

The time it takes to maximum height,
`t_(max)` ≈ 2.0408 seconds

The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height,
`t_(15)` is given as follows;

v₂² = u₂² + 2·g·h₀

v₂² = 20² + 2×9.8×15 = 694

v₂ = √694 ≈ 26.344 m/s

v₂ ≈ 26.344 m/s

From, v₂ = u₂ + g·
`t_(15)`, we have;

26.344 = 20 + 9.8×t

9.8·
`t_(15)` = 26.344 - 20 = 6.344

∴
`t_(15)` = 6.344/9.8 ≈ 0.647

The ball will hit the ground after 2 ×
`t_(max)` +
`t_(15)` ≈ 2 × 2.0408 + 0.647 ≈ 4.7286

The ball will hit the ground after approximately 4.7286 ≈ 4.73 seconds

2. When the ball is thrown upward from the Moon, we have;

The acceleration due to gravity on the moon, a = 1.6 m/s², therefore, we have;

A. The relation between the height, h, and the time, t, after the ball is released is given as follows;

h = h₀ + u·t - 1/2·a·t²

B. The height of the ball after 3 seconds is given by substitution as follows;

At t = 3 seconds, h = 15 + 20 × 3 - 1/2 × 1.6 × 3² = 67.8

The height of the ball, h, thrown on the Moon, after 3 seconds is h = 67.8 m

C. The time the ball takes to hit the ground = 2 × The time it takes to maximum height + The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height

The time it takes to maximum height,
`t_(max)`, is given as follows;

v = u - a·
`t_(max)`

Where;

v = The final velocity = 0 at maximum height

Therefore, we have;

0 = 20 - 1.6 ×
`t_(max)`

∴
`t_(max)` = 20/1.6 = 12.5

The time it takes to maximum height,
`t_(max)` = 12.5 seconds

The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height,
`t_(15)` is given as follows;

v₂² = u₂² + 2·a·h₀

v₂² = 20² + 2×1.6×15 = 448

v₂ = √448 ≈ 21.166 m/s

v₂ ≈ 21.166 m/s

From, v₂ = u₂ + a·
`t_(15)`, we have;

21.166 = 20 + 9.8×t

1.6·
`t_(15)` = 21.166 - 20 = 1.166

∴
`t_(15)` = 1.166/1.6 ≈ 0.72785

The ball will hit the ground after 2 ×
`t_(max)` +
`t_(15)` ≈ 2 × 12.5 + 0.72875 = 25.72875 ≈ 27.73

The ball will hit the ground after approximately 25.73 seconds

3. The height from which the ball is kicked, h₀ = 1 m

The initial velocity of the ball, u = 25 m/sec

The acceleration due to gravity, g = 9.8 m/s²

The relationship between the height, h and the time, t after the ball is released, is given as follows;

h = h₀ + u·t - 1/2·g·t²

B. The height of the ball after 2 seconds is given as follows;

At t = 2, h = 1 + 25 × 2 - 1/2 × 9.81 × 2² = 31.38

The height of the ball, after 2 seconds, h = 31.38 m

C. The time it takes the ball to hit the ground is given by the following kinematic equation, as follows;

h = h₀ + u·t - 1/2·g·t²

At the ground level, h = 0, therefore, we have;

0 = 1 + 25·t - 4.9·t²

Therefore, by the quadratic formula, we have;

t = (-25 ± √(25² - 4×(-4.9)×1))/(2 × -4.9)

Therefore, t ≈ 5.142, or t ≈ -0.03969

Given that the time is a natural number, we have, t ≈ 5.142 seconds

D. The maximum height,
`h_(max)` the ball reaches is given as follows;

From the kinematic equation, v² = u² - 2·g·h,

Where;

v = 0 at maximum height

h = The height the ball reaches above the initial height, we have;

0² = u² - 2·g·h

u² = 2·g·h

h = u²/(2·g) = 25²/(2 × 9.8) ≈ 31.888

`h_(max)` = h₀ + h = 1 + 31.888 ≈ 32.9

The maximum height the ball reaches,
`h_(max)` ≈ 32.9 m