Question

From a stand on top of a bookshelf, you launch a marble with a speed of 3 m/s at a 25 degree angle with the intent of having it land in a small pink bucket that rests on the floor. If the launcher is 1.3 m above the floor. How far from the base of the bookcase should you place the bucket?

Answer 1

Answer:

The distance from the base of the bookcase the bucket should be placed is approximately 2.444 meters

Explanation:

The given parameters are;

The speed with which the marble was launch, u = 3 m/s

The direction in which the marble was launched = 25 degrees

The height of the launcher above the ground, h = 1.3 m

The time it takes the marble to reach maximum height, $t_{max}$, is given as follows;

v × sinθ = g × $t_{max}$

$t_{max}$ = v × sinθ/(g) = 3×sin(25)/9.81 ≈ 0.192 seconds

The time back to the top of the bookshelf level = The time to maximum height  ≈ 0.192 seconds

The time it takes the marble to reach the from top of the bookshelf level is given by the equation for free fall using the vertical component of the velocity as follows;

h = 1/2×g×t²

1.3 = 1/2 × 9.81 × t²

t² = 1.3/(1/2 × 9.81) ≈ 0.265 s²

t ≈ 0.515 seconds

Total time of flight of the marble =  0.192 seconds +  0.192 seconds + 0.515 seconds ≈ 0.899 seconds

The distance from the base of the bookcase the bucket should be placed = The total horizontal range of the marble

The total horizontal range of the marble = The horizontal component of the velocity × Time of flight of the marble

∴ The total horizontal range of the marble = u × cos(θ) × $t_{(Time \ of \ flight)}$ = 3 × cos(25) × 0.899 ≈ 2.444 meters

∴ The distance from the base of the bookcase the bucket should be placed ≈ 2.444 meters.