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, 08:30 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, 10:21 AMGeoC

Plug & Chug for 45 degree takeoff angle

v = 38.6 m/sec - May 27th 2010, 11:15 AMundefined
- May 27th 2010, 12:18 PMGeoC
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 PMtyke
yes, guess I should have said minimum velocity to clear the wall

- May 27th 2010, 03:06 PMtyke
GeoC

in your solution you have

but

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

- May 27th 2010, 04:00 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, 04:16 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, 04:45 AMGeoC
Yes, I have a typo.... should have wrote rather than . However, I did not make this error when I actually worked the problem on paper. My expression for v, i.e.

is correct