1.
x= (90cos60)t, y= 5+(90sin60)t-16t^2
x= 90(.5)t, y= 5+(90(.866))t-16t^2
For horizontal distance t=3 so,
x= 90(.5)3 → x= 135ft
For the height of the ball after 3 seconds,
y= 5+(77.94)3-16(3)^2 → y= 94.82ft
2.
x= (110cos45)t, y= 2+(110sin45)t-16t^2
To find how long the ball...
For calculating the height after 3 seconds, I plugged in the parametric x=(90cos60)t, y=5+(90sin60)t-16t^2 and evaluated for x=3 on the graph, which gave me y= 4.02ft. In addition, when I try to evaluate the horizontal distance by plugging in t=3 I get -257.15
x=(90cos60)t...
I don't know how exactly to find horizontal distance but these are the numbers i came up with after entering the equation in the calculator
x=(90cos60)t y=5+(90sin60)t-16t^2
after 3 seconds: 4.02ft
x=(110cos45)t y=2+(110sin45)t-16t^2
How long the ball will be in the air: 339.273...
Homework Statement
Find the height of the ball after 3 seconds if the ball is hit with an initial velocity of 90ft per second at an angle of 60° from an initial height of 5 ft.
Find the horizontal distance a ball travels after 3 seconds if the ball is hit with an initial velocity of 90 ft...