3. 446 points | Previous Answers GoCP2 4.P020. My Notes Consider a rock that is

Question

3. 446 points | Previous Answers GoCP2 4.P020. My Notes Consider a rock that is thrown off a bridge of height 75 m at an angle = 24 quantities. with respect to the horizontal as shown in the figure below. If the initial speed the rock is thrown is 11 m/s, find the following Bridge (a) The time it takes the rock to reach its maximum height 45 (b) The maximum height reached by the rock. 76.02 (c) The time at which the rock lands. 4.39 d) The place where the rock lands 44.12 (e) The velocity of the rock (magnitude and direction) just before it lands. magnitude 15 direction 55.6 X m/s

Explanation / Answer

(A) v0y = 11 sin24

at max height, vy = 0

vf = vi + a t

0 = 11 sin24 - 9.8 t

t =0.456 sec  

(B) vf^2 - vi^2 =2 ad

0^2 - (11 sin24)^2 = 2(-9.8)(h)

h = 1.02 m

from ground = 75 + h =76.02 m

(C) yf - yi = v0y t + ay t^2 / 2

0 - 75 = (11 sin24) t - 9.8 t^2 /2

4.9 t^2 - 4.474 t - 75 = 0

t = 4.395 sec

(D) x = (v0x) t

= (11 cos24) (4.395)

= 44.17 m

(e) vx = v0x = 10.05 m/s

vy = v0y + ay t = (11 sin24) - (9.8 x 4.395)

vy = - 38.6 m/s

magnitude = sqrt(vx^2 + vy^2) = 39.9 m/s

direction = tan^-1(vy/vx) = - 75.4 deg

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