A softball is hit at a 30 degree angle traveling 100 meters before being caught. At what speed did it leave the bat?

T=2u sin(Angle)÷g should give the flight time ?

33.3 m/s ?

Good point. Wiffle ball and plastic bat would mean less momentum and greater atmospheric friction effects.

Standard density ball and bat would leave the calculations fairly straghtforward.

Would be a speculation based on constants, no an accurate velocity solution.

Unless i just realised that u could solve it using vectors. Using a right angle triangle draw a base line with an gles of 30 degrees draw another line to indicate 100 metres and using trignometry work out the length of the base line and that should give the speed then it just working out the acceleration of the ball after it hit the bat......

We don't even have the elevation where this happened. In what air density? What was the humidity, how much dust in the air, or is that happening in vacuum? Or on the Moon maybe? What was the position of earth in relation of the moon and the sun at that time? Wind direction & speed? And what was the height of the ball at the time the bat hit it? What was the height of the ball at the time of catching up?

We can't solve this if you don't give us proper information about these.

Speed=Distance over Time
S=D/T
So wheres the time?

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Solved

scratch that just realised it dont give the speed when u work out the base line it gives the distance....so back to my original point......

Also, most solutions presuppose a fully elastic collision between the bat and the ball. It never is, and some energy is lost there.

Air resistance does play a role, but the answer would be much different on the moon.

The same speed it was traveling when it was caught if there's no atmosphere.

If you told us what size softball we were using we could use the known mass of the ball, the angle, and distance to calculate the force required to fire it like a cannon. The speed should work itself out from there.

You are so close, that I'm just going to give it to you.

The softball has to leave the bat traveling at 33.6 meters per second / 75.16 miles per hour, in order to travel 100 meters at a 30 degree angle.

Congratulations.

which is why the padres have fleets of moonbats to monitor the winds

If you look up the trajectory of a projectile, you have the angle of the launch and the distance traveled. The time is derived from the gravitational constant as it is averaged through the distance of the arc. It is essentially the height of the arc. The ball is slowed at a constant rate to zero in the up direction and falls in a rate, both directions governed by gravity. The time is thus that which is necessary for a launch at thirty degrees to go 100 meters given that it will lose upward speed/momentum to gravity at a constant rate and gain it going downward also at the rate of gravity. The time is basically twice the time it would take to drop the ball from the top of the arc.

For all we know, a "softball" could be a hardened marshmallow

Insufficient data.

OP failed and dipped.

Lol. Flat earth math. You'd fail trig so hard.

User ID: 55667583 sufficiently answered the question. In long form, the answer is 33.6393597 meters per second.

lol it's magic

My participation in this thread is over. Congratulations User ID: 55667583 for coming extremely close to the right answer of 33.6393597 meters per second.

d = distance = 100m
@ = angle = 30
v = velocity
g = gravity (9.8 m/s^2)

d = (v^2 X sin (2@))/g

v = 56.7 m/s

Retarded OP doth dot
Comprehending significant digits.

No one still got it correct. Even OP is wrong.

100 x Tm / (cos .3)

I love reading mathtard arguments.

45.0 m/s.

Vertical vi = 45 sin 30 = 22.5
horizontal u = 45 cos 30 which is constant

h = hi + vi t - 4.9 t^2

d = u t
100 = 45 cos 30 * t
so
t = (100/45)/cos 30

then
h = 1.3 + 22.5 t - 4.9 t^2

Last Edited by Titcoin on 05/19/2017 08:15 AM