a stone was thrown upward from the top of a cliff overlooking the ocean and its splash was observed 30/7 sec later. The stone's vertical velocity after 1 sec was 21/5 sec. The stone's height in meters is modeled by h(t)= -4.9*t^2 + v0*t + h0, where v0 is the initial velocity, h0 is the initial height and t is times in sec.
a. what was the stone's initial vertical velocity in m/sec?
b. how high is the cliff in meters?
c. what was the stone's maximum height i feet above the ocean
d. what was the stone's velocity in miles per hour at impact?
a. what was the stone's initial vertical velocity in m/sec?
b. how high is the cliff in meters?
c. what was the stone's maximum height i feet above the ocean
d. what was the stone's velocity in miles per hour at impact?
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time in air = 4.2857 sec
V(1) = + or - 4.2 m/sec
height = Ho +Vo*t -1/2*9.8*t^2
H(4.2857) = Ho +Vo*4.2857 -4.9*4.2857^2 = 0
Ho +4.2857Vo = 90
V = V0 -A*t so V = Vo -9.8t
V(1) = + or - 4.2
therefore Vo = +5.6 or +14.0
there are 2 cases to investigate to see if numbers go together
case 1 Vo = +5.6m/s = initial velocity
Ho +4.2857Vo = 90
Ho +24 =90
Ho = 66 = cliff height
set Vo -1/2*a*t^2 =0
5.6 = 4.9*t^2
t^2 =1.2428sec^2 t = 1.0690 sec
this is the time at max height
5.6m/s*1.069s*0.5 = 2.99m max height
Vo -a*t
-5.6 - 9.8 * (4.2857-2*1.069)
-5.6 -9.8*(2.1477) = - 26.647m/s final velocity
V(1) = + or - 4.2 m/sec
height = Ho +Vo*t -1/2*9.8*t^2
H(4.2857) = Ho +Vo*4.2857 -4.9*4.2857^2 = 0
Ho +4.2857Vo = 90
V = V0 -A*t so V = Vo -9.8t
V(1) = + or - 4.2
therefore Vo = +5.6 or +14.0
there are 2 cases to investigate to see if numbers go together
case 1 Vo = +5.6m/s = initial velocity
Ho +4.2857Vo = 90
Ho +24 =90
Ho = 66 = cliff height
set Vo -1/2*a*t^2 =0
5.6 = 4.9*t^2
t^2 =1.2428sec^2 t = 1.0690 sec
this is the time at max height
5.6m/s*1.069s*0.5 = 2.99m max height
Vo -a*t
-5.6 - 9.8 * (4.2857-2*1.069)
-5.6 -9.8*(2.1477) = - 26.647m/s final velocity