The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by a receiver 78 yards directly downfield at a height of 4 feet. The pass is released at an angle of 75 degrees with the horizontal. Find the time in seconds the receiver has to reach the proper position after the quarterback releases the football. Round your answer to one decimal place.

t = 25.1seconds

Step-by-step explanation:

X(t)= 16cos75°t

Y(t)= 6 + (Vosin75°t - 16t^2)

Converting 78 yards to feet :

1 yard = 3 feet

78feet =?

78×3=234 feet

Y(t)= number of feet above the ground at t seconds

VoCos75° t = 234

t = 234/(VoCos75°)

At this time,4 = y(234/VoCos75°)

4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2

Vo= 140.66ft/s

Time,t= 234/(140.66× Cos75°)

t = 25.087seconds

t= 25.1 seconds (1 decimal place)

i need the table so i can see

step-by-step explanation: