Recorded times and positions for the flight of a ball. The quadratic trendline f

Question

Recorded times and positions for the flight of a ball. The quadratic trendline for this y = -5.06t^2 + 2.47 t + 1.2. The distance traveled by the ball during the time interval is: -0.45 m -0.75 m 0.75 m 1. 1.4 m 1.2 m The displacement of the ball during the time interval is: -0.45 m 0.75 m c. 0.75 m d. 1.4 m 1.2 m According to the trendline, the acceleration of the ball during the experiment is: 1.8 m/s^3 b. -10.1 m/s^1 c. 4.9 m/s^2 -4.9m/s^2 -9.8 m/s^2 According to the trendline, the initial velocity at time t = 0.0 s is: -9.8 m/s -4.9 m/s 1.2 m/s d. 2.5 m/s 4.9 m/s The instantaneous speed at time t = 0.25 s is closest to: 9.8 m/s b. 4.9 m/s c. 1.2 m/s d. 1.0 m/s e. 0.1 m/s A ball is released from rest. The expected lime taken by the ball to go through the first 2.5 m distance is: 0.26 s 0.71 s 0.97 s 0.46 s c. 3.0 s A hall is thrown upward with an initial speed of 2.5 m/s. 1 he expected time it takes to teach the top of its motion is: 0.26 s b. 0.71s c. 0.97 s d. 0.46 s c. 3.9 s A ball is thrown upward with an initial speed of 2.5 m/s. What is the expected maximum height of its motion? 0.64 m b. 0.32 m c. 25 m d. 2.0 m e. 9.8 m A ball is released from rest. How far down should it go in a time of 2.5 s? 4 m b. 61 m c. 12 m d. 31 m e. 9.8 m A bull is thrown down with an initial speed of 4.2 m/s. The distance traveled by the ball during the first 2.5 s is expected to be: 41 m b. 11 m c. 26m d. 21 m c. 16m

Explanation / Answer

26.

We use kinematic equation,

h=vi*t+1/2*at^2

Here d=-2.5m, vi=0m/s,a=g=-9.8m/s^2

Put values,

-2.5=0*t-1/2*9.8*t^2

Calculations give t= 0.71s

27.

vi=2.5m/s,vf=0m/s,g=-9.8m/s^2

vf=vi-gt

0=2.5-9.8*t    

Gives, t= 0.26s

28.

vi=2.5m/s,vf=0m/s,g=-9.8m/s^2

vf^2 =vi^2 -2gh

0^2=2.5^2 -2*9.8*h

Gives, h= 0.32m

29.

vi=0m/s, t=2.5s

We use kinematic equation,

h=vi*t+1/2*at^2 = 0*2.5 – ½*9.8*2.5^2 = -30.6 m

30.

vi=4.2m/s, t=2.5s

We use kinematic equation,

h=vi*t+1/2*at^2 = -4.2*2.5 – ½*9.8*2.5^2 = -41.13 m

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