The height from the practice problem is 44m Please help! Due soon PHYSICS % Phy
ID: 2001420 • Letter: T
Question
The height from the practice problem is 44m Please help! Due soonPHYSICS % Phy 101 . Chapter 3-W16 × 22 wekomw/ www web assignnet/webStudent/Assignment-Responses-utrnt 7depe!2921982 e Q Search " EXERCISE HINTS: GETTING STARTEDI rM STUCK Suppose the bal is thrown froth the same height as in the PRACTICE IT problem at an angle of 32.0 below the horizontal. If it strikes the ground 39.7 m away, find the following. (Hist: For part (a), use the equation for the x-dsplacement to eiminate vof from the equation for the y -displacement.) (a) the tme of fight 6 81 The x coordinate as a function of time s x() con(32.0)t, so the initial speed is a-Ax/Ncos 32.0a), whare ax 39.7 and At is the time of fight. Insert this into your equation or y() and solve for the time of fight. Note that the answer should be smaller than 2.997, since the baltl is theown down (and to the right), s tho iritial sped 0.530 Make sure you havefeed coordnatn syster. For rstance, 8 the ongn (0, 0) at the base of the building? If so, then the ritial height of the bal is 44.0 m. 1s the origin nstead at the irtia pocition of the balt? tf so, than the baß larde whan its coordnate ic equal to-440m. s tc) the speed and angle of he velocity vector with respect to the horicona at mpact speed mýs angle o below the horizontal A Sorce Fi of magnitude 5.t0 units acts on an object at the ongin in a drection 0% boe the positve x ais. (See the fgure b bove the positve rans (see the fore b Aforce Flor magnetudo s.BO tnts acts on an obiect at the oge, in a drecton·tky 09"m Cortana. Ask me anything 5 6
Explanation / Answer
v(t) = at + v0
p(t) = 0.5at^2 + v0 * t + p0
Let p0 = (0.0, 43.0)
It lands at position (39.7, 0.0).
Decompose the initial velocity into horizontal (x) and vertical (y) components.
v0x = v0 * cos(-32)
v0y = v0 * sin(-32)
(a) px(t) = 39.7 = 0.5 * 0 * t^2 + v0x * t + 0
39.7 = v0x * t
39.7 = v0 * cos(-32) * t
39.7 / (cos(-32) * t) = v0
py(t) = 0 = 0.5 * -9.8 * t^2 + v0y * t + 43.0
0 = -4.9 * t^2 + v0y * t + 43.0
0 = -4.9 * t^2 + v0 * sin(-32) * t + 43.0
0 = -4.9 * t^2 + (39.7 / (cos(-32) * t)) * sin(-31) * t + 43.0
0 = -4.9 * t^2 + (39.7 / cos(-32)) * sin(-31) + 43.0
0 = -4.9 * t^2 + 39.7 * tan(-32) + 43.0
4.9 * t^2 = 39.7 * tan(-31) + 43.0
t^2 = (39.7 * tan(-32) + 43.0) / 4.9
t = sqrt((39.7 * tan(-32) + 43.0) / 4.9)
t = 1.92s
(b) Substitute t into 39.7/ (cos(-32) * t) = v0 = 24m/s
(c) v(t) = at + v0
vx(t) = 0 * t + v0x
vy(t) = -9.8 * t + v0y
vx(t) = 0 * t + v0 * cos(-32) = 20.35
vy(t) = -9.8 * t + v0 * sin(-32) = 1.537
Use the value of t from (a) and the value of v0 from (b).
The magnitude of v(t) = sqrt(vx(t) ^ 2 + vy(t) ^2). = 20.40N
The angle below the horizontal = -arctan(vy(t) / vx(t)) = 4.31 degree
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.