A ball rolls along a desktop with an uniform velocity of 3 m/s. What is the displacement after 10 seconds have passed?
A ball is thrown upward with an initial velocity of 18 m/s? What is the maximum height the ball reaches if the ball is assumed to leave from the ground?
A ball is dropped from a 35 m tall building. The ball takes how long to impact the ground?
See your 11-8-11, 7:04pm post.
To answer the first question, we need to use the formula for displacement: displacement = velocity × time. Given that the ball has a uniform velocity of 3 m/s, and 10 seconds have passed, we can calculate the displacement by multiplying the velocity (3 m/s) by the time (10 s). Therefore, the displacement after 10 seconds would be 30 meters.
To answer the second question, we need to find the maximum height reached by the ball when thrown upward. This can be done using the kinematic equation for vertical motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * displacement
Since the ball is thrown upward, the acceleration is due to gravity and is equal to -9.8 m/s^2 (negative since it acts opposite to the direction of motion). The initial velocity is given as 18 m/s, and we want to find the displacement at maximum height, which we'll call "h". At maximum height, the final velocity will be 0, as the ball momentarily comes to a stop before falling back down.
Plugging in the values into the equation, we get:
0^2 = (18 m/s)^2 + 2 * (-9.8 m/s^2) * h
Simplifying the equation, we have:
0 = 324 m^2/s^2 - 19.6 m/s^2 * h
Solving for h, we get:
h = 324 m^2/s^2 / 19.6 m/s^2
h = 16.53 meters
Therefore, the maximum height reached by the ball is approximately 16.53 meters.
To answer the third question, we can use the kinematic equation for vertical motion:
displacement = initial velocity * time + (1/2) * acceleration * time^2
The ball is dropped, which means the initial velocity is 0. The acceleration due to gravity is again -9.8 m/s^2, and the displacement is 35 meters (since it is dropped from a 35 m tall building). We want to find the time taken for the ball to impact the ground, so we can rearrange the equation as follows:
displacement = (1/2) * acceleration * time^2
Plugging in the values, we have:
35 m = (1/2) * (-9.8 m/s^2) * time^2
Simplifying the equation, we get:
70 m = -4.9 m/s^2 * time^2
Dividing both sides by -4.9 m/s^2, we have:
time^2 = 70 m / -4.9 m/s^2
time^2 = -14.29 s^2
Since time cannot be negative, we discard the negative sign and take the square root of the result:
time = √(14.29 s^2)
Therefore, the ball takes approximately 3.78 seconds to impact the ground.