A man fires a rock of a slingshot directly upward.The rock has an initial velocity of 15m/s.How long will it take for the rock to return to the level he fired it at?
Use the third formulae of calculating the displacement.ur initial velocity is 15m/s ,ur final velocity is 0m/s,ur displacement is 0m .substitute this values and solve time
To find the time it takes for the rock to return to the level it was fired from, we can use the kinematic equation for displacement:
Δy = V₀t + (1/2)at²
Since the rock is fired directly upward, its initial velocity and final velocity at the top of its trajectory will be zero. Therefore, the equation simplifies to:
0 = V₀t - (1/2)gt²
Where:
V₀ is the initial velocity of the rock (15 m/s)
t is the time taken for the rock to return to the level it was fired from
g is the acceleration due to gravity (approximately 9.8 m/s²)
Rearranging the equation to solve for t:
(1/2)gt² = V₀t
(1/2)gt = V₀
t = (2V₀) / g
Substituting the values:
t = (2 * 15 m/s) / 9.8 m/s²
Evaluating the expression:
t ≈ 3.06 seconds
Therefore, it will take approximately 3.06 seconds for the rock to return to the level it was fired from.
To find the time it takes for the rock to return to the level it was fired, we can use the equation for vertical motion:
h(t) = h0 + v0t + (1/2)gt^2
Where:
h(t) is the height of the rock as a function of time t
h0 is the initial height (the level at which the rock was fired, which we assume to be 0)
v0 is the initial velocity of the rock (15m/s)
g is the acceleration due to gravity (-9.8m/s^2, assuming downward as negative)
t is the time
Given that the initial height is 0, the equation becomes:
h(t) = 0 + (15m/s)t + (1/2)(-9.8m/s^2)t^2
When the rock returns to the level it was fired, the height will be zero. So we can set h(t) to zero and solve for t.
0 + (15m/s)t + (1/2)(-9.8m/s^2)t^2 = 0
To solve this equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
where a = (1/2)(-9.8m/s^2), b = 15m/s, and c = 0.
Plugging the values into the quadratic formula:
t = (-15m/s ± √((15m/s)^2 - 4(1/2)(-9.8m/s^2)(0)))/(2(1/2)(-9.8m/s^2))
Simplifying the equation further:
t = (-15m/s ± √(225m^2/s^2))/(2(-4.9m/s^2))
t = (-15m/s ± 15m/s)/(2(-4.9m/s^2))
Now, solving for t, we have two possible values:
t = (15m/s + 15m/s)/(2(-4.9m/s^2)) ≈ 3.06s
t = (15m/s - 15m/s)/(2(-4.9m/s^2)) ≈ 0s
Since time cannot be negative, the rock will return to the level it was fired at in approximately 3.06 seconds.