A baseball is hit at 40 m/s at an angle of 30 degrees to the horizontal. a) How high will it go? b) When will it reach that height? c) What will be its horizontal distance from the batter at that time?

Question

anonymous · Answer

Someone please help me!

anonymous · Answer

I'm not too into the physics course, our class just started

anonymous · Answer

time to reach the maximum height
use Vf= Vy +gt
        0=20 m/s -9.83t

solve to find t

Vy=v sin 30  (the vertical component of th initial velocity

anonymous · Answer

how high will it go?

use Vf squared =Vy squared + 2 gh

at max height Vf=0 ;  Vy is vertical component of initial velocity v

                         0=40 sin 30   (squared)     + 2(-9.83)     x h
               19.66h= 400   solve to find h

anonymous · Answer

Range= Vx  x time of flight

Vx = Horizontal component of initial velocity=v x Cos 30

time of flight = 2 x time to reach maximum height

hope this has been of some help

anonymous · Answer

you are given the initial velocity of 40m/s and the angle $$\theta = 30$$. this is a projectile motion when you reach the top the final velocity will zero $$v =0$$.  the maximum height that you are looking is the vertical component(y-axis) of the motion then the initial velocity in the y is then$$v _{0}\sin \theta = 40\sin 30 = 20m/s$$. then i would use the equation $$v^{2} = u ^{2} + 2as$$, where $$a = g =  -9.8 m/s ^{?}$$ is the accelaration due to gravity acting downwards and $$s$$ is the height you looking for.$$v^{2} = u ^{2} + 2as \rightarrow s = \frac{ 200 }{ 9.8} m$$