A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? Solution If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With vo= 150 m/s, a = 45°, and g = 9.8 m/s², we have x = 150 cos(z)t = [ y = 10 + 150 sin ¹ (77) - 1/2 (9.8) 6² - X Impact occurs when y=0, that is 4.9t2-75√2t - 100. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t=75√2+√√11,250 + 196 9.8 Then x = 75√2(21.74) 2306.82 x X (rounded to the nearest whole number), so the projectile hits the ground about v(t) = r'(t) = X So its speed at impact (rounded to the nearest whole number) is |v(21.74)| = √(75√2)² + (75√/2 – 9.8 - 21.74)² = ✔ m/s. xm away. The velocity of the projectile is
A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? Solution If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With vo= 150 m/s, a = 45°, and g = 9.8 m/s², we have x = 150 cos(z)t = [ y = 10 + 150 sin ¹ (77) - 1/2 (9.8) 6² - X Impact occurs when y=0, that is 4.9t2-75√2t - 100. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t=75√2+√√11,250 + 196 9.8 Then x = 75√2(21.74) 2306.82 x X (rounded to the nearest whole number), so the projectile hits the ground about v(t) = r'(t) = X So its speed at impact (rounded to the nearest whole number) is |v(21.74)| = √(75√2)² + (75√/2 – 9.8 - 21.74)² = ✔ m/s. xm away. The velocity of the projectile is
Related questions
Question
A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed?
Solution
If we place the origin at ground level, then the initial position of the projectile is
(0, 10)
and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With
v0 = 150 m/s,
? = 45°, and g = 9.8 m/s2, we have| x | = | 150 cos
|
||||||
| y | = | 10 + 150 sin
|
Impact occurs when y = 0, that is
4.9t2 − 75
t − 10 = 0.
Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.)| 2 |
t =
≈
75
|
||||
| 9.8 |
Then
x ≈ 75
(21.74) ≈
(rounded to the nearest whole number), so the projectile hits the ground about m away. The velocity of the projectile is| 2 |
v(t) = r ′(t) =
.
So its speed at impact (rounded to the nearest whole number) is
|v(21.74)| =
≈ m/s.
|
Transcribed Image Text:A projectile is fired with initial speed 150 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed?
Solution
Vo
If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With v
x = 150 cos
(7) ₁-
t =
4
y = 10 + 150 sin
t =
Impact occurs when y = 0, that is 4.9t² - 75√2t - 10 = 0. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.)
75√2 + √11,250 + 196
9.8
(77) ¢ - 1½ (9.8)e² = [
t-
4
2
Then x≈ 75√2(21.74) ≈ 2306.82
v(t) = r'(t) =
X
X (rounded to the nearest whole number), so the projectile hits the ground about
So its speed at impact (rounded to the nearest whole number) is
Iv(21.74)=√(75√2)² + (75√/2 - 9.8 - 21.74)² =
m/s.
m away. The velocity of the projectile is
= 150 m/s, α = 45°, and g
9.8 m/s², we have
Expert Solution
Step 1
Given that the initial speed = 150m/s
Angle Alpha =45°
Initially vertical position= 10 m
g = 9.8m/s2
Step by step
Solved in 3 steps with 2 images
