The water slide ends at height of 1.50m?
First, compute the speed of take-off at the end of the slide using projectile physics x(t)=v0*t 2.5=v0*t the fall time is equal to t y(t)=1.5-.5*g*t^2 when y(t)=0, splashdown solve for t t=sqrt(3/g) plug that into the x equation 2.5*sqrt(g/3)=v0 v0 is related to the loss in PE m*g*h=.5*m*v0^2 solve for h h=.5*v0^2/g plug in v0 h=.5*2.5^2/3 h=1.04 m check v0=4.52 x=2.5 check ====================
Using distance to determine height of a slide?
I'm going to assume the child came off the slide in the horizontal direction so the time to fall 1.8 m is 1.8 = 4.9t^2 t = 0.6 sec. to travel 2.69 m in 0.6 seconds the child came of the slide with a speed of 2.69/.6 = 4.483 m/s v^2 = 2gh 20.1 = 19.6h h = 1.02 meters. Not a tall one .
Determining the height of the water slide... help!?
The child travels L = 2.31 m in the time it takes to fall that 1.80 m under gravity from the end of the slide. Let's take g = 9.81 m/s². We can use one of the standard kinematics equations to find the time s = ut + ½at² where s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time In this case, the shape of the slide means that the child comes off horizontally, with no vertical component of speed, so (u) = 0 m/s 1.80 = 0 + ½ * 9.81 * t² t = 0.606 s The horizontal speed of the child coming off the slide is therefore (2.31 m) / (0.606 s) = 3.81 m/s The slide is frictionless, so energy is conserved. The kinetic energy of the child at the bottom of the slide is equal to the loss in gravitational potential energy coming down the slide. ½mv² = mgh {m cancels out} ½v² = gh v² =2gh h = v²/2g = 3.81² / ( 2 * 9.81) = 0.74 m
Determining the height of a water slide?
At point B, the slide is horizontal. When a child is 1.80 meters above the water, the direction of the child’s velocity is horizontal. At this position, the child’s vertical velocity is 0 m/s. Let’s use the following equation to determine the time for the child to fall 1.80 meters. d = vi * t + ½ * a * t^2, vi = 0, a = 9.8 1.8 = ½ * 9.8 * t^2 t = √(1.8/4.9) This is approximately 0.606 seconds. Let’s use the time and the horizontal distance to determine the child’s velocity at this position. d = v * t 2.12 = v * √(1.8/4.9) v = 2.12 ÷ √(1.8/4.9) = 3.49781544 m/s Use the following equation to determine the vertical distance from this point to the top of the slide. vf^2 = vi^2 + 2 * a * d, vi = 0 vf^2 = 2.12^2 ÷ (1.8/4.9) = 12.3475556 a = 9.8 12.3475556 = 2 * 9.8 * d d = 12.3475556 ÷ 19.6 = 0.6242222 meters To determine the height, add 1.8 meters. This is approximately 2.42 meters
If a block slides down a 30 degree incline at a constant speed, what is the coefficient of dynamic friction between the block and the incline?
θ = 30 degreesvelocity is constant therefore acceleration a = 0Please refer to my analysis using the diagram below.The coefficient of dynamic friction is equal to 0.5773.
An object is placed on an inclined plane 5m high. What is the velocity of object at the bottom of an inclined plane when released from that height?
Gravitation is a conservative force, meaning thereby that the work done in moving between two points is the same irrespective of he path taken between these two points. Whether the body free falls from a height h or slides down a frictionless inclined which at its highest point is having a height h, work done is the same. So potential energy at height h gets converted into kinectic energy equal to the potential energy.Potential energy at height h = m g h.Let body have a velocity v after falling through height h. The kinetic energy at the lowest point= ½ m v². Andm g h = ½ m v², orv = ( 2 g h ) ½.In the present case h = 5 m, taking g = 10 m/s², we get,v = ( 2×10×5)½ = 10 m/s. The body will have the same velocity, whether it has free fall from 5 m height or slides down an incline starting from 5 m height.
Children slide down a frictionless water slide that ends at a height of 1.80 m above the pool.?
If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.33 m from the base of the slide, determine the height h of the water slide. Please help and explain step by step
A body sliding on a smooth inclined plane requires 4sec to reach the bottom after starting from rest at the top. How much time does it take to cover one fourth the distance starting from the top?
A body start from rest so, u=0…Let total distance covered, s…Let a body moves with accerlation, a…s=1/2×a×t×t…s=1/2×a×4×4…s=8a…The one-fourth of total distance,s* =8a/4=2a…s*=1/2×a×t×t…2a/a=1/2 ×t×t… 4=t×t…t=2