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Abc23
An elevator is stopped at the ground floor. It starts moving upwards at constant acceleration a>0 for 5 seconds. It then keeps a constant speed for 35 seconds. Finally, it slows down with an acceleration of the same magnitude (but opposite direction) −a , until it comes to a halt at the top floor. The top floor is 370 meters above the ground floor. (a) What is the maximal speed v of the elevator ? (in m/s)
We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 48 ∘. The gravitational acceleration is g=10 m/s2. (See figure) (a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)
Use the equation for parabola maximum height. Substract from that height the height of wall to get the height of roof. use trignometry tan on the angle beta to get s. If you know how to solve for time, do let me know pls
For the elevator problem to solve its the distance traveled/ 40(this is velocity) for acceleration its the velocity/5 so v=d/40 =370/40 a=v/5
The time to reach the highest point is tₒ= 2vₒ•sinα/2g =2•20•sin35/2•10 =1.15 s The time for covering the distance to the wall ‚d’ is t₁=d/vₒ•cosα =15/20•cos35= 0.92 s At horizontal distance d from the initial point the ball is at the height h₁=vₒ•sinα•t₁ -gt₁²/2 = =20•sin25•0.92 -10•0.92²/2 =6.31 m. The highest position of the ball moving as projectile is H= vₒ²•sin²α/2g =20²• sin²35/2•10 = 6.38 m. Therefore, the ball meets the roof at its upward motion, => d+s=vₒ•cosα•t …..(1) h+s•tanβ= vₒ•sinα•t - gt²/2 …..(2) From (1) s = vₒ•cosα•t -d Substitute ‘s’ in (2) h +tanβ(vₒ•cosα•t –d) = =vₒ•sinα•t - gt²/2, 6+ 0.58(20•0.82•t -15) = 20•0.57•t- 5t², 5t² -1.9t-2.7 =0 t=0.95 s. s= vₒ•cosα•t –d= =20•0.82•0.95 – 15= =15.58 – 15 =0.58 m
370 divided by 35 sec gives you the answer in meters a second