Aquarterback throws a football to a teammate. the football is 6.5ft above the ground when it leaves the quarterback's hand. his teammate catches it 3.5s later, at a height above the ground of 5 ft. determine the value of v, rounded to the nearest whole number.

v=5.42m/s

Step-by-Step Explanation:

We can use conservation of energy to solve this  

mgh=1/2mv^2

2gh=v^2

v=sqrt(2gh)

v=sqrt[2*9.8m/s^2*(6.5-5)]

v=5.42m/s

89

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.