If a robber needed to jump from one window to another, which was 25 m away and 33 m below his own at a 62 degree angle, what would his initial velocity have to be?
Thanks for all the help!
2007-12-03 08:18:47 · 3 answers · asked by chewd.pencils 2 in Science & Mathematics ➔ Physics
The initial velocity is 62 degrees above horizontal?
x(t)=v0*cos(62)*t
y(t)=33+v0*sin(62)*t-.5*g*t^2
when x(t)=25
and y(t)=0 he lands on the window.
25=v0*cos(62)*t
0=33+v0*sin(62)*t-.5*g*t^2
v0*t=25/cos(62)
t=4.04 seconds
v=13.18 m/s
j
2007-12-03 08:28:14 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
Angle 30 Degrees Sin 30 = 0.5 Horizontal velocity = 0.5* 16.4 Vertical Velocity = 16.4 - Horiz vel. Time of flight is time to reach max hight and return to ground Hor displacement = 8.2*t
2016-05-28 00:53:06 · answer #2 · answered by liliana 3 · 0⤊ 0⤋
hay man watch me
when u see a projectile motion u couluse these laws
height=(v(intial)^2 sin(theta)^2)/2g
time=(2v(intial) sin(theta))/g
range(displacement)=(v(intial)^2 sin2(theta))/g
so we have the height and the range so we can use one of them to answer the question
we wil take the height(h)=33m
33=(v(intial)^2*(sin(62)^2))/9.8
33*9.8=(v(intial)^2*0.779596
v(intial)^2=414.8300
v(intial)=20.3673
2007-12-03 08:50:19 · answer #3 · answered by tiger_vision2036 2 · 0⤊ 0⤋