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

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- May 27th 2010, 07:30 AMtykeBaseball 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, 09:21 AMGeoC
$\displaystyle v_x = vcos(\theta ) = constant;$

$\displaystyle v_y = vsin(\theta )t-gt^2/2+1$

$\displaystyle t=d/v_x$

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

Plug & Chug for 45 degree takeoff angle

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

v = 38.6 m/sec - May 27th 2010, 10:15 AMundefined
- May 27th 2010, 11:18 AMGeoC
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, 01:43 PMtyke
yes, guess I should have said minimum velocity to clear the wall

- May 27th 2010, 02:06 PMtyke
GeoC

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

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

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

$\displaystyle 2 = vsin(/theta)t-gt^2/2$ - May 27th 2010, 03:00 PMGeoC
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, 03:16 AMtykegeoc
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, 03:45 AMGeoC
Yes, I have a typo.... should have wrote $\displaystyle y=vsin(\theta)t-gt^2/2+1$ rather than $\displaystyle v_y = .....$. However, I did not make this error when I actually worked the problem on paper. My expression for v, i.e.

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

is correct