Question

A rock is kicked horizontally off a dock into the bay 3

Answer 1

Answer:

The true statement is;

(b) The rock takes about the same time to hit the water in each case

Explanation:

The given information are;

The direction in which the rock is kicked = Horizontally

The number of different initial horizontal with which the rock as kicked =  different horizontal velocity

The time taken for the rock to hit the water is given by the following equation for projectile motion;

y = y₀ + v₀× sin(θ₀)×t - 1/2×g×t²

Where;

θ = The angle the rock makes with the horizontal when it was kicked = 0

y = The height through which the rock drops

y₀ = The initial height of the rock

t = The time taken for the ball to hit the water

g = The acceleration due to gravity ≈ 9.81 m/s²

For the different speeds, v₁, v₂, and v₃, we have;

y = y₀ + v₁× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₂× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

y = y₀ + v₃× sin(0)×t - 1/2×g×t² = y₀ - 1/2×g×t²

Therefore, the time it takes for the ball to hit the water is given as follows;

$t = \sqrt{\dfrac{(y_0 - y) \times 2}{g} }$

Which is the same for each case

However, the speed, v, with which the rock hits the water is given by the following equation;

$v = \sqrt{v_x^2 + v_y^2}$

$v_y$ = The vertical velocity = v₀ × sin(θ₀) - g×t = v₀ × sin(0) - g×t = -g×t

$v_y$ = -g×t

Therefore, $v_y$, is the same for all three times

While vₓ = v₁, v₂, and v₃, for each of the three times.

Therefore, $v = \sqrt{v_x^2 + v_y^2}$  the speed with which the rock hits the water is different for all three times.