(a) Determine the time taken by the projectile to hit point P at ground level.(s

Question

(a) Determine the time taken by the projectile to hit point P at ground level.(s)

(b) Determine the range X of the projectile as measured from the base of the cliff.(km)

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)(m/s)

(d) What is the the magnitude of the velocity?(m/s)

(e) What is the angle made by the velocity vector with the horizontal?(degrees)

(f) Find the maximum height above the cliff top reached by the projectile.(m)

Explanation / Answer

First you need to find the initial velocity components
vx = cos 37 * 105m/s = 83.86m/s
voy = sin 37 * 105 = 63.2m/s

Now to find the time in the air, you use the equation.
y = voy*t + 1/2 gt2

You plug in the variable and use the quadratic equation to solve for time

-225 = 63.2*t + 1/2 *-9.8 *t2

-4.9t2 + 63.2t + 225 = 0

t = -63.2 +/- (63.22 -4*-4.9*225) /2*-4.9

The two solutions that youget are -2.9s and 15.8s and since time cannot be negative, the answer is 15.8s

x = vx * t = 83.86*15.8s = 1325m

You are ready know the horizontal velocity since you are ignoring air resistance (83.86m/s

Vertical velocity = vy = voy + gt = 63.2m/s + -9.8 * 15.8 = -91.64m/s

To find the magnitude, use the pythagorean theorem

91.642 + 83.862 = 124.22m/s

To get the angle tan = opp/adj = tan-1 (91.64/83.86) = 47.54 degrees

To get height:

v2 = vo2 + 2gy

y = 02 - 63.22/2*-9.8 = 203.8m

Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.