3. A marble leaves a 0.70-meter high table with an initial horizontal velocity of 3m / s Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
4. A soccer ball is kicked horizontally off a 20-meter high hill and lands a distance of 32 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
5. A frog leaps horizontally from a 0.30 m high rock with a speed of 0.25 m / s How far from the base of the rock will she land?
4. A soccer ball is kicked horizontally off a 20-meter high hill and lands a distance of 32 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
5. A frog leaps horizontally from a 0.30 m high rock with a speed of 0.25 m / s How far from the base of the rock will she land?
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Answer:
3. We can use kinematic equations to solve for the time and distance:
The vertical motion of the marble is governed by the equation: y = viy*t + 1/2*g*t^2, where y is the vertical distance, viy is the vertical velocity, g is the acceleration due to gravity, and t is the time elapsed.
At the moment the marble leaves the table, its vertical velocity is zero and the vertical distance is 0.70 m. We can assume that the acceleration due to gravity is -9.8 m/s^2.
0.70 m = 0*t + 1/2*(-9.8 m/s^2)*t^2
Solving for t, we get t = 0.38 s.
The horizontal motion of the marble is governed by the equation: x = vix*t, where x is the horizontal distance, vix is the horizontal velocity, and t is the time elapsed.
The initial horizontal velocity of the marble is 3 m/s.
x = vix*t = (3 m/s)(0.38 s) = 1.14 m.
Therefore, the time required for the marble to fall to the ground is 0.38 seconds and the horizontal distance between the table's edge and the ball's landing location is 1.14 meters.
4. We can use kinematic equations to solve for the initial horizontal velocity:
The horizontal motion of the soccer ball is governed by the equation: x = vix*t, where x is the horizontal distance, vix is the initial horizontal velocity, and t is the time elapsed.
The vertical motion of the soccer ball is governed by the equation: y = viy*t + 1/2*g*t^2, where y is the vertical distance, viy is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time elapsed.
At the moment the soccer ball is kicked, its vertical velocity is zero (assuming it is kicked horizontally) and the vertical distance is 20 m.
We can assume that the soccer ball lands at the same height as the hill, so the final vertical distance is zero and the time of flight is the same for the vertical and horizontal motions.
y = viy*t + 1/2*g*t^2
0 = 0*t + 1/2*(-9.8 m/s^2)*t^2 + 20 m
Solving for t, we get t = 2.02 s.
The horizontal distance is given as 32 m.
x = vix*t
Solving for vix, we get vix = x/t = 32 m/2.02 s = 15.8 m/s.
Therefore, the initial horizontal velocity of the soccer ball is 15.8 m/s.
5. We can use kinematic equations to solve for the horizontal distance:
The horizontal motion of the frog is governed by the equation: x = vix*t, where x is the horizontal distance, vix is the initial horizontal velocity, and t is the time elapsed.
We can assume that the acceleration due to gravity is -9.8 m/s^2 and the time of flight is the same for the vertical and horizontal motions.
The vertical motion of the frog is governed by the equation: y = viy*t + 1/2*g*t^2, where y is the vertical distance, viy is the initial vertical velocity, g is the acceleration due to gravity, and t is the time elapsed.
At the moment the frog leaves the rock, its vertical velocity is viy = 0.25 m/s and the vertical distance is 0.30 m.
y = viy*t + 1/2*g*t^2
0.30 m = 0.25 m/s*t + 1/2*(-9.8 m/s^2)*t^2
Solving for t, we get t = 0.338 s.
The horizontal distance is given by:
x = vix*t
Solving for x, we get x = vix*t = (0.25 m/s)(0.338 s) = 0.0845 m or 8.45 cm.
Therefore, the frog will land 8.45 cm from the base of the rock.