A body is thrown downward with aninitial speed of 20 m/s on earth. What is the
A. Acceleration of the object
b. Displacement after 4 seconds
c time required to reach a speed of 50 m/s
i need the answers but also show me how you get them. thanks
2006-11-22 00:27:40 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
ok biker bike, here we go.
acc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)
b) we apply formula
s=ut + 1/2at^2
u=20
t=4
s=20*4 + 1/2 *9.8 *4*4
=158.4 m
c) v=u + at
therefore, t= v-u/a
=50-20/9.8
=3 seconds (approx)
2006-11-22 01:24:38 · answer #1 · answered by dev 1 · 0⤊ 0⤋
a)
The acceleration is acceleration due to gravity called g (9.81ms^-2)
the speed v of the object after time t is : v = gt + v0 (v at time t=0)
to find the distance you must integrate this expression
d = (1/2 ) g t^2 + v0 t +x0 ( we suppose x0 =0)
d = (1/2) 9.81 *16 + 20*4 = 108.8 m
c) use the formula of the speed
v = gt +v0 ---> 50 = 9.81 t +20 t = 30/9.81 =3.3s
2006-11-22 02:50:17 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
i need a time to calculate
2006-11-22 03:24:00 · answer #3 · answered by sara 2 · 0⤊ 0⤋
a) -9.8m/s^2
b) x=(1/2)(-9.8m/s^2)(4s)^2+20m/s(4s)
x=1.6m
c) Vf=at+Vi
-50m/s=(-9.8m/s^2)t+(-20m/s)
-30m/s=(-9.8m/s^2)t
t=3.06s
2006-11-22 01:37:41 · answer #4 · answered by 7 · 0⤊ 0⤋