A. What is the horizontal distance between the catapult and the target? B. How l

Question

A. What is the horizontal distance between the catapult and the target? B. How long did the ball spend in the air before hitting the target? 0 by 9iving thé magnitude and direction of the third leg. oo y giving the C. What was the student's displacement for their walk? 2. (For this problem, please ignore air resistance and assume that the ball always leaves the catapult at the same speed.) A catapult on a level field can toss a spherical 0kg ball a maximum distance of 255m when the catapult is adjusted to the optimal angle. A target is then placed a distance of 20m above the ground and an unknown horizontal distance away from the catapult. The catapult is adjusted to an angle of sixty degrees above the horizontal and is successful at hitting the target. a. What is the horizontal distance between the catapult and the target? b. How long did the ball spend in the air before hitting the target? -4470-94b3-bb38560df596/page/88f6e82d-368c-4d6d-8132-2c232fóc8bdf

Explanation / Answer

Here ,

as maximum range is given as

R = v^2/g

255 = v^2/9.8

v = 50 m/s

Now , for h = 20 m

theta = 60 degree

trajectory of projectile is

y = x * tan(theta) - g * x^2/(2 * (v * cos(theta))^2)

20 = x * tan(60) - 9.8 * x^2/( 2* (50 * cos(60))^2)

solving for x

x = 208.7 m

the target is 208.7 m away .

b)

time of flight = distance/speed

time of flight = 208.7/( 50 * cos(60))

time of flight =8.35 s

the ba;; is in air for 8.35 s

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