a ball is thrown horizontally from the top of a building 35m. high. the ball strikes the ground at a point 80m from the base of the building.
a. find the time the ball is in flight
b. its initial velocity
c. the x and y components of velocity just before it hits the ground
a. = h = 0.5g*t^2 = 35 m.
4.9t^2 = 35.
t^2 = 7.14
t = 2.67 s. Time in flight.
b. Range = Xo*t = 80 m.
Xo*2.67 = 80.
Xo = 30.0 m/s.
c. Y^2 = Yo^2 + 2g*h = 0 + 19.6*35 =
686.
Y = 26.2 m/s.
X = Xo = 30.0 m/s.
Philip jumps horizontally from the top of a building that is twenty and a half meters high. He hopes to reach a swimming pool that is at the bottom of the building, fourteen meters horizontally from the edge of the building. To make it to the pool:
a. The time the ball is in flight can be found using the equation for vertical motion. We can use the formula h = ut + (1/2)gt^2, where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
In this case, the initial vertical velocity is 0 because the ball is thrown horizontally. The height (h) is 35 m, so we can rewrite the equation as 35 = (1/2)(9.8)t^2. Solving for t, we get:
t^2 = (2 * 35) / 9.8
t^2 = 70 / 9.8
t^2 ≈ 7.14
t ≈ √7.14
t ≈ 2.67 seconds
So, the ball is in flight for approximately 2.67 seconds.
b. The initial horizontal velocity can be found using the equation d = vt, where d is the distance and t is the time. In this case, the distance is 80 m, and the time is the same as the time calculated in part (a), which is 2.67 seconds. We can rewrite the equation as 80 = v * 2.67. Solving for v:
v = 80 / 2.67
v ≈ 29.96 m/s
So, the initial velocity of the ball is approximately 29.96 m/s.
c. Just before it hits the ground, the y component of velocity will be 0 since the ball has reached the maximum height and is on its way down. The x component of velocity remains constant throughout the entire flight, and it is equal to the initial horizontal velocity.
Therefore, the x component of velocity just before it hits the ground is approximately 29.96 m/s, and the y component of velocity is 0 m/s.
To solve this problem, we can use the equations of linear motion and the principles of projectile motion. Let's break it down step by step:
a. Finding the time the ball is in flight:
We know that the vertical displacement (Δy) is equal to -35m (negative because the ball is falling downward) and the vertical acceleration (a) is equal to -9.8 m/s² (due to gravity). We can use the equation:
Δy = v₀y * t + (1/2) * a * t²
Substituting in the known values:
-35 = 0 * t + (1/2) * (-9.8) * t²
Simplifying the equation:
-35 = -4.9t²
Rearranging the equation:
4.9t² = 35
Solving for t:
t² = 35 / 4.9
t² = 7.14
t ≈ √7.14
t ≈ 2.67 seconds
So, the ball is in the air for approximately 2.67 seconds.
b. Finding the initial velocity:
We can use the equation:
v₀x = Δx / t
Substituting in the known values:
v₀x = 80 / 2.67
v₀x ≈ 29.96 m/s
So, the initial velocity of the ball in the horizontal direction is approximately 29.96 m/s.
c. Finding the x and y components of velocity just before it hits the ground:
Since the ball is traveling horizontally, the x-component of velocity remains constant. Therefore, the x-component of velocity just before hitting the ground is equal to the initial horizontal velocity:
vxf = v₀x ≈ 29.96 m/s
For the y-component of velocity, we can use the equation:
vyf = v₀y + a * t
Substituting in the known values:
vyf = 0 + (-9.8) * 2.67
vyf ≈ -26.20 m/s
So, the x-component of velocity just before hitting the ground is approximately 29.96 m/s, and the y-component of velocity just before hitting the ground is approximately -26.20 m/s.