Calculus horizontal

Question

An athlete puts a​ 16-lb shot at an angle of 30° to the horizontal from 6ft above the ground at an initial speed of 50 ft/sec.

a. How long is shot in the air

b. How far does shot travel

Answer 1

Solution:

a) Recalling equations for a moving projectile y = -1/2gt² +V_oyt + y_o where y is the final elevation and when the shot is at the ground the final position y = 0, V_oy = 50 ft/s (sin 30) = 25 ft/s, g = 32.2 ft/s², and y_o = 6ft

Then 0 = -1/2 ( 32.2 ft/s²) ( t²) + 25 ft/s ( t ) + 6ft , 0 = -16.1 t² + 25 t +6 rearrange

16.1 t² -25 t - 6 = 0 , solve quadratic equation for t

t = 25 ± √ [( -25)² - 4( 16.1)(-6)] / 2(16.1) = 25 ± √1011.4 / 32.2 = 1.76 s or -0.21 s, the 2nd answer is negative and does not apply , t = 1.76 s

b) the distance the shot travel is found by the following formula

X = V_oxt + X_o here X_o =0 , and

X= 50 ft/s (cos 30) (1.76s) = 76.2 ft