A ball is projected horizontally from the edge of a table that is 1.00 m high, and it strikes the floor at a point 1.30 m from the base of the table.
(a) What is the initial speed of the ball?
m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?
m
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How long does it take to fall 1.00 m?
4.9t^2 = 1
t = 0.45s
(a) 1.3m/0.45s = 2.88 m/s
(b) 2.88 = 9.8t
t = 0.29
h = 1 - 4.9*0.29^2 = 0.58 m
Thank you so much this really helped
(a) To solve for the initial speed of the ball, we can use the fact that the ball was projected horizontally. This means that the vertical component of its initial velocity is zero.
Given:
Height of the table, h = 1.00 m
Horizontal distance from the table to the point where the ball hits the ground, x = 1.30 m
Using the kinematic equation for vertical motion:
y = y_0 + v_0y * t - (1/2) * g * t^2
Since the ball was projected horizontally, there is no vertical displacement, and the equation simplifies to:
0 = 0 + 0 - (1/2) * g * t^2
Rearranging the equation to solve for time, t:
t^2 = (2 * h) / g
t^2 = (2 * 1.00 m) / 9.8 m/s^2
t^2 = 0.2041 s^2
t ≈ 0.452 s
Now, we can use the horizontal distance and time to find the initial speed of the ball:
v = x / t
v = 1.30 m / 0.452 s
v ≈ 2.876 m/s
Therefore, the initial speed of the ball is approximately 2.876 m/s.
(b) To find the height of the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal, we can use the horizontal range formula:
R = v^2 * sin(2θ) / g
Where:
R = horizontal range (1.30 m in this case)
v = initial speed (to be determined)
θ = launch angle (45° in this case)
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation to solve for initial speed, v:
v = sqrt((R * g) / sin(2θ))
v = sqrt((1.30 m * 9.8 m/s^2) / sin(2 * 45°))
v ≈ 8.869 m/s
Now, we can use the initial speed and launch angle to find the height of the ball above the floor using the time-of-flight formula:
h = (v^2 * sin^2(θ)) / (2 * g)
h = (8.869 m/s)^2 * sin^2(45°) / (2 * 9.8 m/s^2)
h ≈ 2.403 m
Therefore, the ball is approximately 2.403 m above the floor when its velocity vector makes a 45.0° angle with the horizontal.
To solve this problem, we need to use the principles of projectile motion and kinematics. Let's go through the steps to find the answers:
(a) What is the initial speed of the ball?
Step 1: Identify known values and variables:
We know that the ball was projected horizontally and hit the floor at a distance of 1.30 m from the base of the table. We can assume that the initial vertical velocity of the ball is zero (since it's projected horizontally) and the acceleration due to gravity is 9.8 m/s².
Step 2: Find the time of flight:
We can use the equation for vertical displacement to find the time of flight. The vertical displacement can be calculated using the equation:
Δy = (1/2) * g * t²
Since the vertical displacement is 1.00 m and the acceleration due to gravity is 9.8 m/s², we can solve for t:
1.00 m = (1/2) * 9.8 m/s² * t²
t² = (1.00 m * 2) / 9.8 m/s²
t² = 0.2041 s²
t ≈ 0.45 s
Step 3: Find the horizontal velocity:
Since the ball is projected horizontally, the horizontal velocity remains constant throughout its motion. We can use the formula:
v = d / t
where v is the velocity, d is the distance, and t is the time. We can plug in the values:
v = 1.30 m / 0.45 s
v ≈ 2.89 m/s
So, the initial speed of the ball is approximately 2.89 m/s.
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?
Step 1: Identify known values and variables:
We want to find the height above the floor when the velocity vector makes a 45.0° angle with the horizontal. We know that the ball was projected horizontally, so the initial vertical velocity is zero, and the acceleration due to gravity is 9.8 m/s².
Step 2: Find the time taken to reach the given angle:
At the given angle, the horizontal and vertical components of velocity have the same magnitude. We can use trigonometry to find the time taken by equating the horizontal distance and vertical displacement.
Using the horizontal velocity (v = 2.89 m/s) and the horizontal distance (d = 1.30 m), we can rearrange the formula:
d = v * t
t = d / v
t ≈ 1.30 m / 2.89 m/s
t ≈ 0.45 s
Step 3: Find the height of the ball:
Using the time calculated in step 2, we can find the height using the equation for vertical displacement:
Δy = v₀y * t + (1/2) * g * t²
Since the initial vertical velocity (v₀y) is zero and the acceleration due to gravity (g) is 9.8 m/s², the equation simplifies to:
Δy = (1/2) * g * t²
Plugging in the values:
Δy = (1/2) * 9.8 m/s² * (0.45 s)²
Δy ≈ 0.993 m
So, the ball is approximately 0.993 m above the floor when its velocity vector makes a 45.0° angle with the horizontal.