SECTION 4 You are outside in a big, level field, in the midst of which stands a
ID: 2307998 • Letter: S
Question
SECTION 4 You are outside in a big, level field, in the midst of which stands a narrow wall 5m high. You are given a device that can launch a projectile, always with the same speed of a 2.0-m/s. (a) First, you launch a projectile from the top of the wall, at an angle of R.8O.5 degrees above the horizontal. How far from the base of the wall does the projectile land on the ground? Answer: m Next, you launch another projectile rom the top of the wall. After s arle flight, the projectile lands on the ground. seconds of (b) At what angle above the horizontal was the projectile launched this time? Answer: degreesExplanation / Answer
to reach the ground, projectile have to travel 15.6m downwards in vertical.
In vertical,
initial velocity, uy = 52sin20.5 = 18.21 m/s
y = - 15.6 m
a = - 9.8 m/s^2
using r = ut + at^2 / 2
-15.6 = 18.21t -4.9t^2
4.9t^2 - 18.21t - 15.6 = 0
t = 4.43 s
distance traveled in horizontal = ux* t
= (52 cos20.5) x 4.43 = 215.77 m
b) from above part, using r = ut + at^2 / 2
in vertical direction,
-15.6 = (52sin@)t - 9.8t^2/2
-15.6 = (52 sin@)6.6 - 4.9(6.6^2)
sin@ = 197.84 / 343.2
@ = 35.20 deg
c) R = vx * t
= (52 cos35.20)6.6 = 280.44 m
d) in horizontal,
speed = 52 cos13.9 = 50.48 m/s
distance = 36.2 m
time = distance / speed = 0.717 s
in vertical,
y = (52 sin13.9)0.717 - 4.9(0.717^2)
y = 6.44 m
e)
in horizontal,
speed = 52 cos@
distance = 36.2 m
distance = speed x time
36.3 = 52t cos@
t = 0.696/cos@
in vertical,
15.6 = (52 sin@)0(0.696/cos@) - 4.9(0.696/cos@)^2
15.6cos^2@ = 36.19 sin@ cos@ - 2.37
15.6 cos^2@ + 2.37 = 36.19 sqrt(1 - cos^2@)cos@
square both sides,
243.36 cos^4 @ + 73.94cos^2 @ + 5.62 = 1309.7 cos^2@ - 1309.71cos^4 @
1553.1 cos^4 @ - 1235.76 cos^2@ + 5.62 = 0
cos^2@ = 0.791 and 0.00457
cos@ = 0.889 and 0.0676
27 deg and 86.12 deg
y = 6.44 m
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