A student club is designing a trebuchet for launching a pumpkin into projectile motion. Based on an analysis of their design, they predict that the trajectory of the launched pumpkin will be parabolic and described by the equation y(x)=ax2+bx where a=−8.0×10−3m−1, b=1.0(unitless), x is the horizontal position along the pumpkin trajectory and y is the vertical position along the trajectory. The students decide to continue their analysis to predict at what position the pumpkin will reach its maximum height and the value of the maximum height. What is the derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position?

The maximum height the pumpkin reaches occurs at 62.5 horizontal meters from its launching spot.

The equation is

that corresponds to a curve presented by a parabola.

Now we can determine the maximum of this parabola with arms pointing down requesting the derivative slope of the tangent line to the curve to be zero

So,

= 62.5m

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The maximum height the pumpkin reaches occurs at 62.5 horizontal meters from its launching spot.

Explanation:

Notice that we are given the actual trajectory equation:

which corresponds to a curve represented by a parabola.

We can find the maximum of this parabola with arms pointing down requesting the derivative (slope of the tangent line to the curve) to be zero :

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s