a ball is projected horizontally from the edge of a table that is 0.433m high, and it strikes the floor at a point 1.84m from the base of the table. the acceleration of gravity is 9.8m/s^2.what is the initial speed of the ball? how high is the ball above the floor when its velocity vector makes a -18.9486 angle with the horizontal?
i got initial speed that is 6.13 m/s. but i don't know to find another question.. please help...
The time it takes to hit the ground is
t = sqrt(2H/g) = 0.2973 s
The speed when it leaves is the constant horizontal component
Vx = 1.84/0.306 = 6.19 m/s
When the velocity vector angle is -18.9486 degrees below horizontal, the ratio Vy/Vx is the tangent of 18.9486 degrees, which is 0.3433
Therefore Vy = 2.13 m/s at that time.
The time after leaving is
t' = 2.13/g = 0.217 s
Use that time to calculate the distance it has fallen, and from that, the height above the floor.
hey i tried bt i didn't got the answer.. can u help more to get answer.. please
To find the initial speed of the ball, we can use the equations of motion for projectile motion. In this case, the ball is projected horizontally, so its initial vertical velocity is zero.
Let's break down the given information:
- The height of the table (h) = 0.433 m
- The horizontal distance from the base of the table to the point where the ball strikes the floor (x) = 1.84 m
- The acceleration due to gravity (g) = 9.8 m/s²
To find the initial speed (v₀), we can use the following equation for the vertical displacement of a projectile:
h = (1/2)gt²,
where t is the time taken for the ball to fall from the table to the floor.
First, let's find the time (t):
Since the ball is projected horizontally, the time taken to reach the floor is the same as the time taken for the ball to fall freely from the table to the floor. We can use the equation for vertical displacement to find this time:
h = (1/2)gt²
0.433 = (1/2)(9.8)(t²)
0.433 = 4.9t²
t² = 0.433 / 4.9
t² = 0.0884
t ≈ √0.0884
t ≈ 0.2976 s
Now that we have the time (t), we can find the initial speed (v₀) by using the equation for the horizontal displacement (x):
x = v₀t
1.84 = v₀(0.2976)
v₀ = 1.84 / 0.2976
v₀ ≈ 6.17 m/s
So, the initial speed of the ball is approximately 6.17 m/s.
Now, let's move on to the next question: how high is the ball above the floor when its velocity vector makes a -18.9486° angle with the horizontal?
We can use the equations of motion to solve this question as well. Since the velocity vector makes an angle, we need to split the velocity into horizontal and vertical components.
Since the ball was projected horizontally, the horizontal component of velocity (v₀x) remains constant throughout the motion and is equal to the initial speed (v₀) we just calculated.
The vertical component of velocity (v₀y) is given by:
v₀y = v₀ * sin(θ),
where θ is the angle (-18.9486° in this case).
Plugging in the values:
v₀y = 6.17 * sin(-18.9486°)
v₀y ≈ -1.99 m/s (negative because it is directed downwards)
Now, we can find the time taken (t') for the ball to reach this height using the equation for vertical displacement:
y = v₀yt' + (1/2)gt'²,
where y is the vertical displacement from the floor.
Let's assume y is the height of the ball above the floor.
Since the ball starts from the floor, the initial vertical displacement is 0. Therefore, the equation becomes:
y = (1/2)gt'²,
y = (1/2)(9.8)(t')².
Now, let's substitute the known values:
0.433 = (1/2)(9.8)(t')².
Simplifying further:
0.433 = 4.9(t')²,
(t')² = 0.433 / 4.9,
(t')² ≈ 0.0884,
t' ≈ √0.0884,
t' ≈ 0.2976 s.
So, the time taken for the ball to reach the given height is approximately 0.2976 seconds.
Now, we can find the height (y) using the equation:
y = v₀yt' + (1/2)gt'².
Plugging in the values:
y = (-1.99)(0.2976) + (1/2)(9.8)(0.2976)²,
y = -0.5924 + 0.0441,
y ≈ -0.5483 m.
The height of the ball above the floor when its velocity vector makes a -18.9486° angle with the horizontal is approximately 0.5483 m (or 54.83 cm).
Please note that the height is negative because the ball is below the point of projection.