Question

Two students are on the roof of a building 29.4 m above the street. One student throws a rock vertically downward at 4.9 m/s. At the same time, the other student throws a ball vertically upward at 4.9 m/s. The second ball misses the roof on its way down.
a) Calculate the difference in the time the rocks spend in the air.
b) Calculate the velocity of each rock as it strikes the ground.
c) Calculate the distance between the rocks 0.5 seconds after they are thrown.

Answer 1

The motion of an object that is due to only gravity is a falling free fall motion

The values of the motion of the rocks are;

• (a) The difference in the time the rocks spends in the air, is 1 second

• (b) The velocity of each rock as it strikes the ground, is 24.5 m/s each

• (c) The distance between the rocks 0.5 seconds after they are thrown is 4.9 meters

The reason the above values are correct are as follows;

The known parameters are;

The height from which the two students throw the rock, h₀ = 29.4 m

Direction and speed with which one throws the rock, u₁ = 4.9 m/s, downwards

Direction and speed with which the second throws the rock, u₂ = 4.9 m/s, upwards

Required:

(a) The difference in the time each rock spends in the air;

Solution:

The kinematic equation for distance travelled, s, is given as follows;

$h = h_0 + u\cdot t - \dfrac{1}{2} \cdot g \cdot t^2$

When the rock reaches the ground, h = 0, we get:

For the rock thrown downwards

$0 = 29.4 - 4.9\cdot t - \dfrac{1}{2} \times 9.8 \times t^2$

0 = 29.4 - 4.9·t - 4.9·t²

0 = 4.9×(6 - t - t²) = -4.9×(t² + t - 6)  = -4.9×((t + 3)·(t - 2))

Therefore, t₁ = -3, or 2

Therefore, the time it takes the rock to land is 2 seconds

For the rock thrown upwards

$0 = 29.4 + 4.9\cdot t - \dfrac{1}{2} \times 9.8 \times t^2$

0 = 29.4 + 4.9·t - 4.9·t²

0 = 4.9×(6 + t - t²) = -4.9×(t² - t - 6)  = -4.9×((t - 3)·(t + 2))

Therefore, t₂ = 3, or -2

Therefore, the time it takes the rock to land is 3 seconds

The difference in the time the rocks spends in the air, is t₂ - t₁

t₂ - t₁ = 3 - 2 = 1 seconds

The difference in the time the rocks spends in the air, is 1 second

b) To calculate the velocity of each rock as it strikes the ground

Solution:

Taking downward motion as positive, we have;

v = u + g·t

For the rock thrown downwards

v₁ = 4.9 + 9.8×2 = 24.5

The velocity of each rock as it strikes the ground, v₁ = 24.5 m/s

For the rock thrown upwards

v₂ = -4.9 + 9.8×3 = 24.5

The velocity of each rock as it strikes the ground, v₂ = 24.5 m/s

c) The distance between the rocks 0.5 seconds after they are thrown

Solution:

The location of the rock thrown downwards 0.5 second after it is thrown is given as follows;

$s_{(1, \, 0.5)} = 29.4 - 4.9\times 0.5 - \dfrac{1}{2} \times 9.8 \times 0.5^2 = 25.725$

The rock thrown downwards is 25.725 m above the ground, 0.5 seconds after it is thrown

The location of the rock thrown upwards 0.5 second after it is thrown is given as follows;

$s_{(2, \, 0.5)} = 29.4 + 4.9\times 0.5 - \dfrac{1}{2} \times 9.8 \times 0.5^2 = 30.625$

The rock thrown upwards is 30.625 m above the ground, 0.5 seconds after it is thrown

The distance between the rocks is $s_{(2, \, 0.5)} - s_{(1, \, 0.5)} =30.625 - 25.725 = 4.9$

Therefore, the distance between the rocks 0.5 seconds after they are thrown is 4.9 meters

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