A ball is thrown horizontally from a window 27 m above the ground. The ball strikes the ground at a point 80 m from the base of the building. The acceleration due to gravity is 9.8 m/s2. What is the resultant velocity of the ball just before it strikes the ground?
37.76 m/s
25.72 m/s
41.12 m/s
53.41 m/s
52.24 m/s
V^2 = vo^2 + 2g*h.
V^2 = 0 + 19.6*27, V = 23 m/s.
To find the resultant velocity of the ball just before it strikes the ground, we need to use the equations of motion. Since the ball is thrown horizontally, we know that the initial vertical velocity (u_y) is 0 m/s.
First, let's find the time (t) it takes for the ball to reach the ground. We can use the equation:
h = u_y * t + (1/2) * g * t^2
Where h is the vertical distance (27 m) and g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation, we get:
(1/2) * g * t^2 = h
Substituting the values, we have:
(1/2) * 9.8 * t^2 = 27
Multiplying both sides by 2 and dividing by 9.8, we get:
t^2 = (2*27) / 9.8
t^2 = 5.51
Taking the square root of both sides, we find:
t ≈ 2.35 s (approximately)
Now that we know the time it takes for the ball to reach the ground, we can find the horizontal velocity (u_x) using the equation:
u_x = d / t
Where d is the horizontal distance (80 m) and t is the time found earlier.
Plugging in the values, we have:
u_x = 80 / 2.35
u_x ≈ 34.04 m/s (approximately)
Since the ball is only affected by gravity in the vertical direction, the vertical velocity (v_y) just before it strikes the ground will be the same as the initial velocity (u_y), which is 0 m/s.
Now, we can find the resultant velocity (v) using the Pythagorean theorem:
v = √(u_x^2 + v_y^2)
Plugging in the values, we have:
v = √(34.04^2 + 0^2)
v ≈ 34.04 m/s (approximately)
Therefore, the resultant velocity of the ball just before it strikes the ground is approximately 34.04 m/s.