16) A cat leaps to try to catch a bird. If the cat\'s jump was at 60° off the gr
ID: 1580741 • Letter: 1
Question
16) A cat leaps to try to catch a bird. If the cat's jump was at 60° off the ground and its initial velocity was 2.74 m/s, what is the highest point of its trajectory, neglecting air resistance? A) 0.29 m B) 0.58 m C) 10.96 m D) 0.19 m Answer: A 0 17) A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g 9.81 m/s2. A) 202 m/s B) 9.80 m/s C) 19.6 m/s D) 45.2 m/s E) 94.4Explanation / Answer
along vertical
initial velocity voy = 2.74 m/s
acceleration ay = -g = -9.8 m/s^2
initial position y0 = 0
final position y = h
final velocity at max height vy = 0
vy^2 - voy^2 = 2*ay*(y - y0)
0^2 - 2.74^2 = -2*9.8*(h-0)
h = 0.384m <<<-------ANSWER
====================
17)
PROJECTILE
1)
along horizontal
________________
initial velocity v0x = v
acceleration ax = 0
initial position = xo = 0
final position = x = 100
displacement = x - x0
from equation of motion
x - x0 = v0x*T+ 0.5*ax*T^2
x - x0 = v*costheta*T
T = (x - x0)/(v*)......(1)
along vertical
______________
initial velocity v0y = 0
acceleration ay = -g = -9.8 m/s^2
initial position y0 = h
final position y = 0
from equation of motion
y-y0 = v0y*T + 0.5*ay*T^2 .........(2)
using 1 in 2
y-y0 = (v*sin(theta)*(x-x0))/(v*cos(theta)) - (0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2)
y-y0 = 0 - ((0.5*g*(x-x0)^2)/v^2)
0-24 = 0 - 0.5*9.8*100^2/v^2
v = 45.2 m/s <<<----------ANSWER
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.