# Math Help - Baseball homerun problem

1. ## 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

2. $v_x = vcos(\theta ) = constant;$
$v_y = vsin(\theta )t-gt^2/2+1$
$t=d/v_x$

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

Plug & Chug for 45 degree takeoff angle

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

v = 38.6 m/sec

3. 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.

4. 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).

5. yes, guess I should have said minimum velocity to clear the wall

6. 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:

$2 = vsin(/theta)t-gt^2/2$

7. 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?

8. ## 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

9. 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