Question

A grasshopper jumps straight up from the ground with an initial vertical velocity of 8 feet per second. a. Write an equation that gives the height (in feet) of the grasshopper as a function of the time (in seconds) since it leaves the ground. b. After how many seconds is the grasshopper 1 foot off the ground?

Answer 1

Answer:

The grasshopper will be at 1 foot after 1.50 seconds.

Step-by-step explanation:

a). A grasshopper jumps straight up with an initial vertical velocity = 8 feet per second

From the equations of vertical motion,

$h=ut-\frac{1}{2}gt^{2}$

Here h = height of the grasshopper at the time 't'

u = initial velocity

g = Acceleration due to gravity

t = duration or time

Now we plug in the value of initial velocity in this equation.

$h=8t-\frac{1}{2}gt^{2}$

b). In this part we have to calculate the time when the grasshopper is 1 foot off the ground.

from the expression of part (a),

$1=8t-\frac{1}{2}gt^{2}$

$2=16t-(9.8)t^{2}$

$-(9.8)t^{2}+16t-2=0$

t = $\frac{-16\pm \sqrt{(16)^{2}-4(-9.8)(-2)}}{2(-9.8)}$

 = $\frac{-16\pm \sqrt{(16)^{2}-78.4}}{2(-9.8)}$

 = $\frac{-16\pm 13.32}{-19.6}$

 = 1.50, (-0.14) seconds

But the time can not be negative, so the grasshopper will be at 1 foot after 1.50 seconds.