here,
the mass of snowball , m = 1.1 kg
the height of cliff , h0 = 14.8 m
initial velocity , v0 = 18.5 m/s
theta = 33 degree
a)
let the speed at the bottom of the ground be v
using conservation of energy
0.5 * m * v0^2 + m * g * h0 = 0.5 * m * v^2
0.5 * 18.5^2 + 9.81 * 14.8 = 0.5 * v^2
solving for v
v = 25.2 m/s
the speed of the snowball at the bottom is 25.2 m/s
b)
when the launch angle is below the horizontal
let the speed at the bottom of the ground be v
using conservation of energy
0.5 * m * v0^2 + m * g * h0 = 0.5 * m * v^2
0.5 * 18.5^2 + 9.81 * 14.8 = 0.5 * v^2
solving for v
v = 25.2 m/s
the speed of the snowball at the bottom is 25.2 m/s
c)
the speed does not depends upon the mass of ball
so, the speed will be the same .i.e 25.2 m/s
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Halliday, Fundamentals of Physics, 10e Grossmont College Physics (PHYC-140-240-24 Chapter 03, Problem 015 [ Incorrect. The two vectors a b in the figure have and equal magnitudes of 14.3 m and the angles are 8,- 340 and 6, - 106%. Find (a) the x component and (b) the y component af their vector sum , (o) the magnitude of F, and (d) the angle 7 makes with the m and the a direction of the x axis. umnber (b) Number...
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Pls answer with solutions Return to Blackboard PLUS Halliday, Fundamentals of Physics, 10e Help I System Announcements i Unread) PRINTER VERSION 4 ВАСН SOURCES NEXT Chapter 02, Problem 052 A bolt is dropped from a oblem o the valley below the bridge. (a) How much time does it take to pass through the last 18 % of its fall? What is its speed (b) when it begins that last f its fall and (c) just before it reaches the or...