ball drops 45m under g=10m/s/s
45=1/2x10xt^2 ... application of kinematic equaion from rest
90/10=t^2
t=3
24.0 m in 3 secs => 8m/s no air resistance
The ball's initial speed is calculated using the principles of projectile motion. First, the time it takes for the ball to hit the ground is found using the vertical distance and acceleration due to gravity. the initial speed to be approximately 7.9 m/s.
The problem describes a case of projectile motion, a common topic in physics. Since the ball is thrown horizontally, the initial vertical velocity of the ball is zero. We're given that the horizontal distance covered is 24.0 m and the vertical distance is 45.0 m.
Because the horizontal and vertical motions are independent, we can use the equations of motion to solve the problem. First, we have to find the time it takes for the ball to hit the ground. Using the equation of motion
"y = 0.5*g*t²",
where y = 45 m is the vertical distance, g = 9.8 m/s² is the acceleration due to gravity, and t is the time in seconds. Solving for t gives us the square root of (2*y/g), which is approximately 3.03 seconds.
Second, we use this time to find the initial speed of the ball. The horizontal distance covered x = 24.0 m is equal to the product of the time it's been travelling and its initial horizontal speed (v = x/t). Using the time from the previous step, we can find the initial speed to be approximately 7.9 m/s.
#SPJ2
The acceleration of gravity is 9.81 m/s?.
If the upward acceleration of the bucket is
2.8 m/s², find the force exerted by the rope
on the bucket of water.
1. 69.658
2. 52.84
3. 85.318
4. 70.86
5. 84.429
Explanation:
Draw a free body diagram of the bucket. There are two forces:
Tension force T pulling up.
Weight force mg pulling down.
Sum of forces in the y direction:
∑F = ma
T − mg = ma
T = mg + ma
T = m (g + a)
Given m = 4.0 kg, g = 9.81 m/s², and a = 2.8 m/s²:
T = (4.0 kg) (9.81 m/s² + 2.8 m/s²)
T = 50.44 N