Range of a Projectile We saw in Problem 56 of Exercises 4.2 that if a projec- tile, such as a shot put, is released from a height h, upward at an angle 0 with velocity v, the range R at which it strikes the ground is given by , cose Vo sin@ + Vvž sin°o + 2gh ). R- where g is the acceleration due to gravity. See FIGURE 4.6.4. (a) Show that when h = 0 the range of the projectile is vở sin 20 (b) It can be shown that the maximum range Ru, is achieved when the angle 0 satisfies the equation gh vở + gh" cos 20 Show that maximum range is Vvô+ 2gh R

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Chapter1: Functions And Models
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16.
75. Range of a Projectile We saw in Problem 56 of Exercises 4.2 that if a projec-
tile, such as a shot put, is released from a height h, upward at an angle 0 with
velocity v, the range R at which it strikes the ground is given by
V, cose
R
Vo sin@ + Vvž sin²0 + 2gh
where g is the acceleration due to gravity. See FIGURE 4.6.4.
(a) Show that whenh = 0 the range of the projectile is
Ground
ở sin 20
FIGURE 4.6.4 Projectile in
Problem 75
(b) It can be shown that the maximum range R, is achieved when the angle 0
satisfies the equation
gh
+ gh
cos 20
Show that maximum range is
vô + 2gh
by using the expressions for R and cos 20 and the half-angle formulas for the
sine and the cosine with / = 20.
Transcribed Image Text:16. 75. Range of a Projectile We saw in Problem 56 of Exercises 4.2 that if a projec- tile, such as a shot put, is released from a height h, upward at an angle 0 with velocity v, the range R at which it strikes the ground is given by V, cose R Vo sin@ + Vvž sin²0 + 2gh where g is the acceleration due to gravity. See FIGURE 4.6.4. (a) Show that whenh = 0 the range of the projectile is Ground ở sin 20 FIGURE 4.6.4 Projectile in Problem 75 (b) It can be shown that the maximum range R, is achieved when the angle 0 satisfies the equation gh + gh cos 20 Show that maximum range is vô + 2gh by using the expressions for R and cos 20 and the half-angle formulas for the sine and the cosine with / = 20.
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