HYS 2110-Fall18 Activities and Due Dates HW: Motion in a Straight Line 9/21/2018

Question

HYS 2110-Fall18 Activities and Due Dates HW: Motion in a Straight Line 9/21/2018 10:59 PM 80.2/100 9/6/2018 03:23 PM B Gradet Print Calculator . Perodic Table Question 30 of 33 Map deh Sapling Learning You throw a baseball directly upward at time t0 at an initial speed of 12.5 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? ignore air resistance and take g·9.80 ms' Maximum height Number Earlier time at half maximum height: Number Later time at half maximum height: OPrevous Give Up & View Souion Check Answer Next Ext earch #00 VVNNA

Explanation / Answer

(a) Intial velocity of the baseball, u = 12.5 m/s

At the maximum height, velocity of the baseball, v = 0

suppose, maximum height attained by the baseball = h meter.

use the expression -

v^2 = u^2 + 2*a*h

put the values -

0 = (12.5 m/s)^2 + 2*(-9.8 m/s^2)*h

=> h = (12.5 m/s)^2 / (2*9.8 m/s^2) = 7.97 m

(b) Half the maximum height = h/2 = 7.97/2 = 3.98 m

use the expression to find out the time to cover this distance -

h = u*t + (1/2)*g*t^2

put the values -

3.98 = 12.5t + 0.5*(-9.8)*t^2 = 12.5t - 4.9t^2

=> 4.9t^2 - 12.5t + 3.98 = 0

t = [12.5 + sqrt(12.5^2 - 4*4.9*3.98)] (2*4.9) =  [12.5 + 8.84] (9.8) = 2.18 s

Second value of t = [12.5 - sqrt(12.5^2 - 4*4.9*3.98)] (2*4.9) =  [12.5 - 8.84] (9.8) = 0.37 s

Therefore, enter time at half maximum height = 0.37 s

And later time at half maximum height = 2.18 s.

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