Hello, Jenny!
Sorry, your friend is right.
. . You forgot about the 20-foot tower . . .
. . x .= .70t
. . . . . . . . . . . . ._
. . y .= .20 + 70√3t - 16tē
HI ,
Could you please help me to check my answer for the below question? Thanks
A projectile is launch with an initial speed of 140feet per second from a 20 feet tall tower at an angle of pi/3 to the horizontal. Assume that the only force acting on the object is gravity.
Find the maximum horizontal range.
Solution:
x=70t
y=70sqrt(3)t -16t^2
The projectile will reaches its maximum horizontal range when it hit the ground that is when y=0.
y=70sqrt(3)t -16t^2=0
so t=0 (ignored) or t= 35sqrt(3)/8
x=70t =70 x 35sqrt(3)/8 = 530.44 ft
So the maximum horizontal range is 530.44ft.
Is my answer correct? The reason I asked that because my friend got a different answer. Her answer is 541.8feet.