it?dep. 18008320 % https// w wet assignet 51 Homework 19 Se- et Student Assignme
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it?dep. 18008320 % https// w wet assignet 51 Homework 19 Se- et Student Assignment. Responses/sub View Favorites Tools Help plingMATLAB Math 151 Engineering Mat., Math 151Cengage A ALEKS District on Luther HOWDY G Goog -2 points If a ball is thrown vertically upward from the ground with a velocity of 96 n/s, then its height after t seconde is s-96e-16e What is the maximum height reached by the ball? What is the velocity of the ball when it is 128 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 128 ft above the ground on its way down? ft/s When will the ball hit the ground? With what velocity does the ball hit the ground? ft/s The position function of a particle is given below st- r - 100, t2 0 When does the reach a velocity of S m/s? to a law of motion s -tt), t 2 0, where t is measured in seconds and s in feet. (a) Find the velocity at time t t) O Type here to search DELExplanation / Answer
Solution:
2) Maximum height
=> s = 96 t - 16 t^2
=> ds/dt = 96 - 32 t
=> at maximum height ds/dt =0
so, 0 = 96 - 32 t
=> t = 3 sec
So maximum height s = 96 x3 - 16 x (3)^2
=> S = 144 ft
b) 128 = 96t - 16t^2
=> 16t^2 - 96 t + 128
=> t^2 - 6t + 8 = 0
=> t^2 - 4t - 2t +8 =0
=> t = 4 , 2 seconds
velocity of the ball on its way up = 96 - 16 x 2 = 64 ft /s
velocity of the ball on its way down = 96 - 16 x4 = 32 ft/s
the ball will hit the ground t = 6 seconds
the ball will hit the ground with velocity = - 96 ft/sec ( if negative sign doesn't work please try with 96 ft/sec as quesiton is not stating about the sign )
Q.3) Given s(t) = t^3 + 3t^2 -100t
=> v = d s / dt = 3 t^2 + 6t - 100
=> 5 = 3 t^2 + 6t - 100
=> 0 = 3t^2 + 6t -105
=> t^2 + 2t - 35 = 0
=> t^2 +7t - 5t -35 = 0
=> t = -7 , 5
time should not be negative so t = 5 seconds
4) Given s = t^3 - 15 t^2 + 72 t
velocity = ds / dt = 3t^2 - 30t + 72
v = 3t^2 - 30t + 72
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