Physics... A football is kicked at an angle of 45 degrees above horizontal.?

Question

a) compute for maximum height
b) total time of flight if it covers a distance of 150ft
c) initial velocity of the ball

Need help.

Answer 1

The horizontal and vertical components of the velocity V are
Vh=Vcos(45)
Vv=Vsin(45)

a)The max height H is
H=0.5gt^2 whre time t
t=Vv/g
and g- acceleration due to gravity

b)Total time is distance over horizontal component of the velocity
t(total)= S/Vh


c) I'm afrain we have made a full circle
V=Vh/cos(45)

We need one more bit more information

Answer 2

V0y = v0 sin theta v0 sin 45 =
V0y = 0.5V0
V0x = v0 cosine theta
V0x =0.5V0

Since the range is given = 150, then
X = V0^2 sin 2 theta/g = v0^2*1/9.8
150*9.8 = V0^2
V0 = 38.34 feet/second. This is the initial velocity of the ball
Hence, V0x = V0y = 19.17 feet/second
T peak occurs when Vy = 0
Vy = V0y –gt; when Vy = 0, then V0y =9.8*t
19.17/9.8 =t; t= 1.96 seconds
T peak = 1.96 seconds, therefore h or Y max height =:

H= Ymax = v0yt -0.5*g*t^2 = 37.57-18.82 =18.75 feet.

Total time of flight = 2* t peak = 3.92 seconds