projectile is fired in the air from a height of 125 feet. Its velocity (in feet per second) t seconds after launch is f(1), given in the following table. 10 15 20 25 30 f(x) 232 209 194 170 148 122 108 a) How fast is the projactile traveling 15 seconds after launch? feet per second. b) How far does the projectile move between 5 seconds and 25 seconds? Estimato toet. c) How tar does the projectile move between 0 seconds and 15 seconds? Estimate feet. di How high is the projectile at 25 seconds? t is foet Of necessary, round to two decimal places)
projectile is fired in the air from a height of 125 feet. Its velocity (in feet per second) t seconds after launch is f(1), given in the following table. 10 15 20 25 30 f(x) 232 209 194 170 148 122 108 a) How fast is the projactile traveling 15 seconds after launch? feet per second. b) How far does the projectile move between 5 seconds and 25 seconds? Estimato toet. c) How tar does the projectile move between 0 seconds and 15 seconds? Estimate feet. di How high is the projectile at 25 seconds? t is foet Of necessary, round to two decimal places)
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Transcribed Image Text:projectile is fired in the air from a helght of 125 feet. Its velocity (in feet per second) t seconds after launch is f(t), given in the following table.
15
10
15
20
25
30
f(x)
232
209
194
170
148
122
108
a) How fast is the projectile traveling 15 seconds after launch? feet per second,
b) How far does the projectile move between 5 seconds and 25 seconds?
Estimato
foet
c) How far does the projectile move between 0 seconds and 15 seconds?
Estimate
feet.
di How high is the projectile at 25 seconds? t is foot
0f necessary, round to two decimal places)

Transcribed Image Text:a) From the giving data, the speed is 160 ft/s
b) Let's calculate the acceleration from a graph
v vs t
250
200
150
100
y = -4.1786x + 231.54
50
10
15
20
25
30
35
t(s)
Remember that:
v = vo + at
Then, the acceleration is the slope:
a = -4.1786
The vertical position is giving by:
1
y = yo + vot +5at?
At t= 5.00 s
1
Vs = 125 + 233 + 5 +(-4.1786)(5)?
Ys = 1237.7675 ft
At t= 25 s.
1
Yas = 125 + 233 + 25 +(-4.1786) (25)?
Y25 = 4644.1875ft
Then, the distance traveled is:
Ay = y2s - Ys = 3406.42 ft
c) Att= 15 s.
Ay = vạt +zat
1
Ay = (233)(15) +;(-4.1786)(15)2
Ay = 3025 ft
d) The vertical position at t = 25 s.
Y25 = 4644.2 ft
(s/1)A
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