A boy throws a rock with an initial velocity of 2.47 m/s at 30.0 degrees above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory?
then divide by 2 to get the hight at time.
Just do the vertical problem
Vi = 2.47 sin 30 = 1.23 m/s
v = Vi - 4.9 t
at top, v = 0
0 = 1.23 - 4.9 t
solve for t
To find the time it takes for the rock to reach the maximum height of its trajectory, we can use the kinematic equation for vertical motion. The equation is:
vf = vi + at
Where:
vf is the final velocity
vi is the initial velocity
a is the acceleration
t is the time
At the maximum height, the final velocity becomes zero since the rock momentarily stops moving before it starts falling downwards.
So, we can rewrite the equation as:
0 = vi + a*t
Since the initial velocity is given as 2.47 m/s and we know that the acceleration due to gravity is -9.8 m/s^2 (since it's acting in the opposite direction), we can substitute the values into the equation:
0 = 2.47 + (-9.8)*t
Now, we can solve for t:
-2.47 = -9.8*t
Dividing both sides by -9.8:
t = -2.47 / -9.8
t ≈ 0.252 seconds
Therefore, it takes approximately 0.252 seconds for the rock to reach the maximum height of its trajectory.
To find out how long it takes for the rock to reach the maximum height of its trajectory, we need to analyze the vertical motion separately. The horizontal motion will not affect the time it takes for the rock to reach its maximum height.
First, we need to break down the initial velocity into its vertical and horizontal components.
The vertical component (v_y) can be found using the formula:
v_y = v * sin(θ)
where v is the magnitude of the initial velocity (2.47 m/s) and θ is the angle of projection (30.0 degrees).
v_y = 2.47 m/s * sin(30.0 degrees) = 1.235 m/s
The time it takes for the rock to reach maximum height can be found using the following equations:
v_fy = v_iy + a * t
v_fy = 0 (at maximum height, the final vertical velocity is 0)
v_iy = 1.235 m/s (vertical component of initial velocity)
a = -9.8 m/s^2 (acceleration due to gravity, which acts in the opposite direction to the motion)
0 = 1.235 m/s - 9.8 m/s^2 * t
Solving for t:
9.8 m/s^2 * t = 1.235 m/s
t = 1.235 m/s / 9.8 m/s^2
t ≈ 0.126 seconds
Therefore, it takes approximately 0.126 seconds for the rock to reach the maximum height of its trajectory.