fjdd The velocity of a projectile fired up into the air follows the equation sho
ID: 1329657 • Letter: F
Question
fjdd The velocity of a projectile fired up into the air follows the equation shown below: v(t) = 40 - 4 t m/s Determine the time when the projectile reaches its maximum height For the velocity in Problem #1, determine the values of time when the velocity is larger than 4 m/s. (i.e., v ( t ) > 4 ) For the velocity in Problem #1, determine the values of time when the magnitude of the velocity is less than or equal to 8 m/s. (i.e., v(t) Ge 8) For the velocity in Problem #1, determine the values of time when the magnitude of the velocity exceeds 12 m/s. (i.e., v (t) > 12)Explanation / Answer
(1)
the given equation is
v (t) = 40 - 4 t
at t = 0 the initial velocity will be
u = v (0)
= 40 m / s
the maximum height attained will be
Hmax= u2/ 2 g
= (40 x 40) (2 x 9.8)
= 81.6 m
time taken to reach maximum height will be the time of ascent
ta = 40 / 9.8
= 4.08 s
(2)
if the velocity greater than 4 that is
v(t) > 4 say it is 5 m / s then
5 = 40 - 4 t
t = 35 / 4
= 8.75 s
(3)
when v(t) = 8.0 m / s
8.0 = 40 - 4 t
4 t = 40 - 8
t = 32 / 4
= 8 s
(4)
if exceeds 12 m / s then let the velocity be 13.0 m / s then
13.0 = 40 - 4 t
t = 6.75 s
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