a brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal and with an initial speed of 15 m/s.
what is the velocity of the brick at the max height? (the answer is 13.6 but I don't understand why or how to do it) (i thought it would be 0m/s)
velocity at max height? horizontal velocity is 15*cosine(25)
the velocity in the vertical direction is of course, zero at the top.
ikr but my teach gave me the answers and it says 13.6m/s
To find the velocity of the brick at the maximum height, we first need to determine the vertical and horizontal components of the initial velocity.
Given:
- Initial speed (Vi) = 15 m/s
- Angle of projection (θ) = 25 degrees
Step 1: Finding the vertical component of the initial velocity (Vy):
Vy = Vi * sin(θ)
In this case:
Vy = 15 * sin(25)
Vy ≈ 6.38 m/s (rounded to two decimal places)
Step 2: Finding the time taken for the brick to reach the maximum height (t):
We know that the time taken to reach the maximum height is the same as the time taken to return to the ground. The vertical motion of the brick can be modeled using the equation:
h = Vy * t - (1/2) * g * t^2
Since the brick reaches the maximum height, the final displacement (h) is zero. So we have:
0 = Vy * t - (1/2) * g * t^2
Step 3: Solve for time (t):
Rearrange the equation:
(1/2) * g * t^2 - Vy * t = 0
Since we are looking for the positive time (the time taken to reach the maximum height), we can simplify the equation by dividing both sides by t:
(1/2) * g * t - Vy = 0
Solve for t:
(1/2) * g * t = Vy
t = (2 * Vy) / g
Given that the acceleration due to gravity (g) is approximately 9.8 m/s^2, substitute the values:
t = (2 * 6.38) / 9.8
t ≈ 1.03s (rounded to two decimal places)
Step 4: Finding the horizontal component of the initial velocity (Vx):
Vx = Vi * cos(θ)
In this case:
Vx = 15 * cos(25)
Vx ≈ 13.57 m/s (rounded to two decimal places)
Step 5: Finding the velocity at the maximum height (V):
The velocity at the maximum height is the horizontal component of the initial velocity (Vx) since there is no horizontal acceleration involved. Therefore, V = Vx.
So, the velocity of the brick at the maximum height is approximately 13.57 m/s (rounded to two decimal places).
The answer you provided, 13.6 m/s, is slightly rounded from the calculated value.