Question

7. A projectile is fired from point O at the edge of a cliff with an initial speed v0 = 187 m/s. The initial horizontal velocity component is v0x = 175 m/s. The projectile rises and then falls into the sea at point P at distance D = 2660 m from the bottom of the cliff. Ignore air resistance.

1a. What is the projectile’s speed 6.73 s after the launch?

b. What is the height of the cliff?

c. What is the launch angle?

Answer 1

we can find total time in air using horizontal motion equation

x = v_ox * t

t = x / v_ox

t = 2660 / 175

t = 15.2 sec

and

we can find the angle of launch from

v_ox = v_o cos

cos = 175 / 187

= 20.637 degree

so,

y_f = y_o + v_yt - 1/2gt²

0 = y_o + 187 * sin 20.637 * 15.2 - 4.9 * 15.2²

y_o = 130.27 m

and

v_fy = v_oy - gt

v_fy = 187 * sin 20.637 - 9.8 * 6.73

v_fy = - 0.04658 m/s

so,

v = sqrt ( (- 0.04658)² + 175²)

v = 175 m/s

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a) What is the projectile’s speed 6.73 s after the launch ?

v = 175 m/s

b) What is the height of the cliff ?

h = 130.27 m

c)  What is the launch angle ?

= 20.637 degree