A startled armadillo leaps upward, rising .544 m in the first .200 s. (a) what is its initial speed as it leaves the ground? (b) What is its speed at the height of .544 m? (c) How much higher does it go?
The answers are (a)3.7 m/s (b) 1.74 m/s (c) .154 m
Can you please show me the steps involved with reaching these answers?
first u should know this formula:X2=1/2at^2+V1t+X1
here a is “-g” because armadillo is going up but g is downward.also we assume that X1 is earth surface,so it’s 0
then the formula changes to:X2= -1/2gt^2+V1t
now we got all the parameters except V1:
X2=.544 ,t=.200 ,g=9.8
.544=1/2×9.8x(.2^2)+.2xV1…so: .544=.196+.2xV1,then V1=.74/.2=3.7m/s
for finding V2 u just need to know that V2= -gt+V1
again we put minus for g because g is downward but the motion is upward.so:…V2= -9.8x.2+3.7…….so:V2=1.74m/s
and for finding How much higher does it go,first u should find the total amount of it’s height,then subtract .544 from it
do u know this formula?(V2)^2-(V1)^2= -2gX
V2 should be 0 here.because in the peak of it’s motion,it doesn’t have any speed.and again for the same reason we put a minus for g.
then wa got:0-(3.7)^2= -2×9.8xX,so:X=13.69/19.6=.698
so the last thing:X2-X1=.698-.544=.154m
hope i helped.:)
persian
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