Another question about projectiles launched at an angle. A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0m/s. If the ball was caught 42.0m from the thrower, how long was it in the air? How high above the thrower did the ball travel?

Question

anonymous · Answer

$$t=\frac{ 42 }{ 23\cos25 }$$

anonymous · Answer

y is now the height of the tower$$y=t(Vsintheta -4.9t)$$

anonymous · Answer

where did you get the -4.9?

anonymous · Answer

it's 1/2 * a
since a is -g
and g is 9.8
acceleration in the y direction is -g for projectile
then 1/2 of -9.8 = -4.9

anonymous · Answer

Okay thanks. duh haha.

anonymous · Answer

good lol

anonymous · Answer

http://www.bgcs.k12.oh.us/Portals/19/Science_HS/Hover/Useful%20Formulas.jpg

anonymous · Answer

so basically use Vy=vi*sin25-1/2gt   ???

anonymous · Answer

yea
but remember it's vi*sin25*t-1/2*g*t^2

anonymous · Answer

i got a physics test tomorrow so i'm practicing by answering question

anonymous · Answer

same here I'm totally flipping out

anonymous · Answer

im doing college phsyics , harder stuff lol

anonymous · Answer

ugh
I didn't get the right answer...it's supposed to be 4.8 m
I got the seconds just not the Dy yet

anonymous · Answer

and I got -8.4m
o_o

anonymous · Answer

dy is 42
we assume the projectile is launched from the origin of our coordinate system

anonymous · Answer

so wait. what am I solving for when it asks how high above the thrower did the ball travel?

anonymous · Answer

vi is v initial
so h=(vi^2*sin^2thetha)/2g
that's to find height, which is gonna be the max height of the parabola

anonymous · Answer

|dw:1350958550651:dw| just to be clear, is this drawing sort of right?
for a normal diagram I guess

anonymous · Answer

|dw:1350958844927:dw|

anonymous · Answer

still not getting the height :/