A catapult on a cliff launches a large round rock towards a ship on the ocean be

Question

A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H of 33.0m above sea level, directed at an angle above the horizontal with an unknown speed v_0. The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 166m. Assuming that air friction can be neglected, calculate the value of the angle. (Answer in "deg".) Calculate the speed at which the rock is launched. To what height above sea level does the rock rise?

Explanation / Answer

height above sea level = 33 m

horizontal distance travelled = 166 m

time = 6 sec

let the initial velocity be v

then horizontal velocity = v * cos(theta)

speed = distance / time

v * cos(theta) = 166 / 6

v * cos(theta) = 27.67 m/s ---------------(1)

vertical speed = -v * sin(theta)

by second equation of motion

s = ut + 0.5 * at^2

33 = -v * sin(theta) * 6 + 0.5 * 9.8 * 6^2

v * sin(theta) = 23.9 m/s -------------(2)

on dividing 2 and 1 we'll get

v * sin(theta) / v * cos(theta) = 23.9 / 27.67

tan(theta) = 23.9 / 27.67

theta = 40.82 degree

value of angle = 40.82 degree

v * sin(40.82) = 23.9

v = 36.56 m/s

launching speed of the rock = 36.56 m/s

by third equation of motion

v^2 = u^2 + 2as

0 = (v * sin(theta))^2 - 2 * 9.8 * s

0 = 23.9^2 - 2 * 9.8 * s

s = 29.143 m

total distance above sea level = 33 + 29.143

total distance above sea level = 62.143 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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