A punter kicks a football. Its height, h(t), in metres, t seconds after the kick is given by the equationh (t)=−4.9t2+18.24t+0.8. The height of an approaching blocker’s hands is modelled by the equationg (t)=−1.43t+4.26, using the same t. Can the blocker knock down the punt.
Please show full solution

Yes

Step-by-step explanation:

−4.9t² + 18.24t + 0.8 = −1.43t + 4.26

-4.9t² + 19.67t - 3.46 = 0

4.9t² - 19.67t + 3.46 = 0

Using quadratic formula:

t = [19.67 +/- sqrt(319.0929l)]/9.8

t = (19.67 +/- 17.8631716109)/9.8

t = 3.8299154705, 0.1843702438

Since the two can be at the same height at the same time, the blocker can knock down the punt

the answer to the expression is 83

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47