A boy stands on the edge of a cliff of height 60 m...?

Question

He throws a stone vertically upwards so that its distance, h m, above the cliff top is given by h=20t-5t^2.

(a) Calculate the time which elapses before the stone hits the beach (it just misses the boy and the cliff on the way down)
(b)Calculate the speed with which the stone hits the beach.

Who can help me with this maths problem?

Answer 1

when the stone hitsthe beach h=-60
-60=20t-5t^2
=>5t^2-20t-60=0
5t^2-30t+10t-60=0
5t(t-6)+10(t-6)=0
(t-6)(5t+10)=0
ignoring the negative value t=6 secs
dh/dt=20-10t
=20-60
=-40 m/sec
the negative sign indicates the downward direction

Answer 2

For an uniformly accelerated movemnt:
space = space(t=0) + speed(t=0)*t + (1/2)acceleration*t^2

In the way upwards:
h = 20t -5t^2 --> space(t=0) = 0, speed(t=0)=20 and acceleration=-5.

(I) Maximun attained heigh is given by the condition speed=0. So:
speed(t(hmax)) = dh/dt = 20-10t
dh/dt=0 --> 20-10t=0 --> t=5secs.

(II) The stone takes again 5 secs to come back to the boy, and the
speed is equal to the initial speed but in opposite direction: -20.

(III) From that moment, the equation of the movement is:
h=-20T-5T^2, considering T=0 when the stone is in h=0 in its way downwards. The time needed to arrive to h=-60m is:
-60=-20T-5T^2
5T^2+20T-60=0 or T^2+4T-12=0
T=(-4+sqrt(4^2+4*1*12))/2*1 = 4 seconds

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So:
(a) Total elapsed time = 5 + 5 + 4 = 14secs.
(b) Speed = dh/dT (T=4) = -20 -10T = -20 -10*4= -60 m/sec.

Answer 3

calculate t such that h = -60 m

-60 = 20t-5t^2
5t^2-20t+60=0
t^2-4t+12=0
(t-6)(t+2)=0
so t is either 6 time units or -2 time units. (There is no such thing as negative time)

change in distance/time = velocity

20-10t at t = 6 means
20-60 = -40 m/time

this means the stone is traveling 40meters/time unit in the downward direction

Answer 4

He's in slightly higher than normal gravity.
a)
-60 = 20t - 5t^2
t^2 -4t - 12 = 0
(t - 6)(t + 2) = 0
t = 6 sec

b)
v = dh/dt = 20 - 10t
v = 20 - 10*6
v = -40 m/sec, or 40 m/s downward

Answer 5

pay attention! people who say that the two stones hit the floor on the precise comparable time are incorrect. they could be wright if the two stones have been brought to the comparable time with the comparable velocity, which isn't the case: whilst the stone you thrown interior the air passes back with the help of you, it has a definite velocity (the only which you gave once you threw it) while the different one has no velocity. a is a persevering with (the only stress exerced on thestones is their weight so a=g=9.81ms^-2) and a=dv/dt so v=-at+Vo ( -a because of fact acceleration is oriented in direction of the graoud it relatively is the alternative orientation of the vertical axes)(Vo is the rate at t=0), and v=dx/dt so x=-0.5at^2+Vot+Xo (x is the peak of the stone, Xo is the peak of the object at t=0) Xo=50 and a=9.80 one so x=-0.5*9.80 one*t^2+Vot+50 shall we determine the preliminary velocity of the 1st stone at t=0 we throw the stone. at t=3.4 we've x=Xo so Xo=-0.5*9.80 one*3.4^2+Vo*3.4+Xo so -0.5*9.80 one*3.4^2+Vo*3.4=0 so Vo=sixteen.7m/s now shall we determine the time that each and each ball needs to hit the floor (this suggests whilst x=0, and we take t=0 the 2d as quickly as we drop the 2d stone) for the 1st stone: Vo=sixteen.7 x=0 x=-0.5*9.80 one*t^2+Vot+50 so 0=-0.5*9.80 one*t^2+sixteen.7t+50 we've 2 values for t: t=5.3s or t=-a million.9s it relatively is impossible so the 1st stone hits the groud after 5.3s for the 2d stone: Vo=0 x=0 x=-0.5*9.80 one*t^2+50 so 0=-0.5*9.80 one*t^2+50 so t=3.2s the 2d ball needs 3.2s to hit the floor shall we substract the two values to determine what number seconds aside do the two stones strike the floor: 5.3-3.2=2.1s

Answer 6

hmm Mechanics eh.

Remember the formulae:
v=u+at
v^2=u^2 + 2as
s=ut+0.5*at^2
s=(t(u+v))/2

Then write down what info you can collect from the stone falling:
s (distance)=
u (initial speed)=
v (final speed)=
a (acceleration)=
t(time)=

Then use an appropriate formula to solve

Answer 7

120 MPH