1.) A projectile is launched with an initial speed of 40.0 m/s at an angle of 30
ID: 583706 • Letter: 1
Question
1.) A projectile is launched with an initial speed of 40.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectile's velocity at the highest point of its trajectory? ___ m/s (b) What is the straight-line distance from where the projectile was launched to where it hits its target? (Note that the hill may slope up or down from the launch point.) ___ m 2.)A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 6.80 m high, to form a 1.1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. ___ m/s (b) Find the vertical distance by which the ball clears the wall. ___ m (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. ___ m
Explanation / Answer
1)
a) 40*cos30=34.64m/s
b)34.64*4=138.56 m
2)
set the equation of motion in the vertical as y=0 at the surface of the playground
for the ball
y(t)=-5.7v0*sin(53)*t-.5*9.81*t^2
we are given that
24=v0*cos(53)*2.20
solve for v0
18.12 m/s
check with y(t) to see if the ball is above the wall
y(2.2)=-5.7+18.12*sin(53)*2.2-.5*9.81*2...
y(2.2)=4.762 m, which is 3.762 m above the wall.
Let's find t for when y(t)=0
There will be two roots, use the greater root. The smaller value of t is when the ball reaches the plane of the playground on the ascent.
t=2.753 seconds
In that time the ball has traveled
x(2.753)=35.5 m
or
35.5-26 from the wall
9.5 m
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