...the kick...? A place kicker must kick a football from a point 39.0 m from a goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked the ball leaves the ground with a speed of 21.6 m/s at an angle of 53° to the horizontal.
(a) By how much does the ball clear or fall short of clearing the crossbar?
Answer in meters.
..................
I got a maximum height for the football at 15.3 using
y = (Vyi)(t) - (.5)(g)(t)^2
Vyi= Velocity of y initial
t = time
g = gravity (9.8 m/s^2)
Ok, nevermind on what I did.
....
I did another calculation:
T=D/Vx
T=39/13 = 3seconds
Then I used:
Y= (Vyi)(t) - (.5)(g)(t)
When I plugged everything in, I get 3.9m and subtracted 3.05m to get .85m. The answer I got is apparently wrong.
(a) By how much does the ball clear or fall short of clearing the crossbar?
Answer in meters.
..................
I got a maximum height for the football at 15.3 using
y = (Vyi)(t) - (.5)(g)(t)^2
Vyi= Velocity of y initial
t = time
g = gravity (9.8 m/s^2)
Ok, nevermind on what I did.
....
I did another calculation:
T=D/Vx
T=39/13 = 3seconds
Then I used:
Y= (Vyi)(t) - (.5)(g)(t)
When I plugged everything in, I get 3.9m and subtracted 3.05m to get .85m. The answer I got is apparently wrong.