Question

A team is building a ballistic ball launcher. The target is 18 ft above the ground, and it needs to catch the ball at the top of its trajectory. Your launcher throws balls from 0.50 m above the ground and must be located 6 m from the target. At what speed must the launcher toss the food in m/s?​ At what angle above the horizontal must the launcher toss the food?​ Explain.

Answer 1

`\theta = 58.98`°

Step-by-step explanation:

given data:

h = 18 ft = 5.48 m

from figure

`h_(max) = 5.48 - 0.50 = 4.98 m`

`h_(max) = ( v_y^2)/(2g)`

`v_y =\sqrt{2gh_(max)`

`v_y = √(2*9.8* 4.98) m/s`

`v_y = 9.88 m/s`

`t = (v_y)/(g) =(9.88)/(9.8) = 1.01 s`

`v_x = (d)/(t) = (6)/(1.01)  = 5.49 m/s</p><p>[tex]v = √(v_x^2+v_y^2) = √(5.95^2+9.88^2)`

v = 11.53 m/s

`tan\theta = \frac[v_x}{v_y}`

`\theta = tan^(-1) (9.88)/(5.94)`

`\theta = 58.98`°