Hugo lies on top of a building, throwing pennies straight down to the street below. The formula for the height, H, that a penny falls is H=Vt+5t2H=Vt+5t2, where V/is the original velocity o f the penny (how fast Hugo throws it when it leaves his hand) and t is equal to the time it takes to hit the ground. The building is 60 meters high, and Hugo throws the penny down at an initial speed of 20 meters per second. How long does it take for the penny to hit the ground?
A. 2/5 seconds
B. 1 second
C. 2 seconds
D. 5 seconds
E. 6 seconds
OA is c
what is the approach to use to determine the accurate answer? Thanks for your prompt response
Algebra
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Hi Roland2rule,Hugo lies on top of a building, throwing pennies straight down to the street below. The formula for the height, H, that a penny falls is H=Vt+5t2H=Vt+5t2, where V/is the original velocity o f the penny (how fast Hugo throws it when it leaves his hand) and t is equal to the time it takes to hit the ground. The building is 60 meters high, and Hugo throws the penny down at an initial speed of 20 meters per second. How long does it take for the penny to hit the ground?
A. 2/5 seconds
B. 1 second
C. 2 seconds
D. 5 seconds
E. 6 seconds
OA is c
Let's take a look at your question.
The formula for the height H, that a penny falls is given by:
$$H=vt+5t^2$$
The building is 60m high, It means
$$H=60m$$
Initial speed of penny is given by 20m/s, therefore,
$$v=20\ \frac{m}{s}$$
We are asked to find the time 't' the penny takes to hit the ground.
Let's plugin all the given values in the formula of height.
$$60=20t+5t^2$$
$$5t^2+20t-60=0$$
Factor out 5:
$$5\left(t^2+4t-12\right)=0$$
$$t^2+4t-12=0$$ $$t^2+6t-2t-12=0$$
$$t\left(t+6\right)-2\left(t+6\right)=0$$
$$\left(t-2\right)\left(t+6\right)=0$$
$$\left(t-2\right)=0\ or\ \left(t+6\right)=0$$
$$t=2,\ -6$$
Since time can not be zero, therefore we will eliminate t = -6.
Therefore, the penny will hit the ground in 2 seconds.
Option C is correct.
Hope it helps.
I am available if you'd like any follow up.
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