Question

During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60^o with the horizontal. If the point of landing is 20 ft away, determine the approximate speed at which the bike was traveling just before it left the ground. Neglect the size of the bike for the calculation.

Answer 1

Answer:

Velocity is equal to $27.3$ feet per second  and time is equal to $1.4668$ seconds

Explanation:

Given

The horizontal distance traveled by  dirt bike before landing $= 20$ feet

Angle of flight $= 60$ degree

As we know that Horizontal distance (H) is equal to

$= H_0 + V_0 * t$

Where $H_0$ is the initial horizontal distance

$V_0$ is the velocity with which the bike is travelling in horizontal direction

and $t$ is the time in seconds

Substituting the given values, we get -

$H = H_0 + v*t\\20 = 0 + v * cos \theta * t\\20 = v * cos 60 * t\\t = \frac{20}{v * cos 60} \\t = \frac{40}{v}$

Now distance traveled in vertical direction is equal to

$Y = Y_0 + v_0 * t + \frac{1}{2} a * t^2$

here acceleration will be equal to acceleration due to gravity which is equal to $- 32.2 \frac{ft}{s^2}$. It is negative as its is acting in upward direction

Thus,

$Y = 0 + v * sin 60 + \frac{1}{2} * (-32.2) * (\frac{40}{v} )^2\\0 = 0 + 0.866 v + \frac{-25760}{v^2} \\0.866 v = \frac{-25760}{v^2}\\v^3 = \frac{-25760}{0.866} \\v = 27.3$

Velocity is equal to $27.3$ feet per second  and time is equal to $1.4668$ seconds