# Baseball homerun problem

• May 27th 2010, 08:30 AM
tyke
Baseball homerun problem
If a baseball leaves the bat at an angle of 45degrees 1m above the ground and clears a wall 3m high 150m away from home plate, what initial velocity would the ball have had if air resistance is negligible
• May 27th 2010, 10:21 AM
GeoC
$v_x = vcos(\theta ) = constant;$
$v_y = vsin(\theta )t-gt^2/2+1$
$t=d/v_x$

$v=\sqrt{\frac{\frac{gd^2}{2cos^2(\theta)}}{d\tan\t heta -2}}$

Plug & Chug for 45 degree takeoff angle

$v=\sqrt{\frac{gd^2}{d-2}}$

v = 38.6 m/sec
• May 27th 2010, 11:15 AM
undefined
Quote:

Originally Posted by tyke
If a baseball leaves the bat at an angle of 45degrees 1m above the ground and clears a wall 3m high 150m away from home plate, what initial velocity would the ball have had if air resistance is negligible

We need to know how high the baseball is when it clears the wall.

I haven't checked GeoC's answer/work, but I'm assuming it's for when the baseball just clears the wall, thus it is a lower bound for initial velocity.
• May 27th 2010, 12:18 PM
GeoC
That's right.. I assumed y=3' at the wall.... I worked through the problem very quickly, so wouldn't be surprised if I've made an error... (although the units on the formula derived for V are correct).
• May 27th 2010, 02:43 PM
tyke
yes, guess I should have said minimum velocity to clear the wall
• May 27th 2010, 03:06 PM
tyke
GeoC

in your solution you have $v-y=vsin(/theta)t-gt^2/2/2+1$

but $v-y=vsin(/theta)$

the height at wall compared to height at release is 2, so from equations of uniform motion:

$2 = vsin(/theta)t-gt^2/2$
• May 27th 2010, 04:00 PM
GeoC
you say "but....." implying I've made a mistake, yet you haven't convinced me that is the case. What do you believe is the actual answer?
• May 31st 2010, 04:16 AM
tyke
geoc
v-y = vsin(theta)

the equation you appear to have used is the standard equation of uniform motion s=ut+0.5at^2. your use of this equation would imply s= vy-1
• May 31st 2010, 04:45 AM
GeoC
Yes, I have a typo.... should have wrote $y=vsin(\theta)t-gt^2/2+1$ rather than $v_y = .....$. However, I did not make this error when I actually worked the problem on paper. My expression for v, i.e.

$v=\sqrt{\frac{gd^2}{d-2}}$

is correct