What horizontal range of the baseball?

Question

What horizontal range of a baseball that leaves the bat at an angle of 63 degree with the horizontal with an initial velocity of I60 km/h. Disregard air resistance.

Answer 1

Watch out, there's a gotcha in this problem. The velocity is given in km/hr, so we have to convert g = 9.8 m/s into km/hr: 9.8 meters = 0.0098 km and 1 second = 1/3600 hr = 0.000278 hr, so g = 0.0098 / 0.000278 = 35.25 km/hr

Horizontal velocity = v * cos(63) = 160 * 0.454 = 72.6 kph.

Vertical velocity = v * sin(63) = v * 0.891 = 142 kph.

The time to attain maximum height is the same time it would take if dropped from that height: 142 kph = g * t, so t = 142 / 35.25 = 4.03 seconds = 0.00112 hr.

This maximum height occurs at exactly halfway through the flight, and since the horizontal velocity is assumed to be constant, the range = 72.6 * 0.00112 * 2 = 0.163 km = 163 meters.

If 10.0 m/s^2 is used for g, the final answer wil be different by a small amount.

Answer 2

This sounds like a baseball game. Hmmm.

The complexity of the problem is the ball (assuming it is still in the stadium) will be hit higher than the it will be hit the ground. Lets now assume that it will be hit 1.5 meters higher by the batter than it will strike the ground.

OK, resolve the vertical and horizontal speeds.
Vvert = Vo*Sin(63); Vhorz = Vo * cos(63).

You do the actual math.

The vertical height will be gotten from the the simple equation;
V(final) = V(initial) -9.8m/s * Time.
(this assumes it on the surface of the earth where the accretion is 9.8m/s.)

The vertical speed will be = 0 at the max height. So,
V(final) = 0 = v(initial) - 9.8m/s * time or time = Vinitia/9.8

The height will be; h=1/2 g *time^2.

The time it takes to drop will be slightly longer since it will have to fall 1.5 m farther. or Time(fall) = sqrt(2*(h+1.5)/g)

The time to fly up and to drop is the time of flight.

The horizontal distance the ball will travel will be the time of flight times the horizontal speed, or ;
D= Time-of-flight * Vh

Now if a fielder catches the ball at the same height as the batter hit the ball the equations get a bit simpler.

Distance = 2* Vo/g * sin(angel) * cos(angle)

Answer 3

contained in the vertical body: initial speed is 0. 40 = a million/2 gt^2. hence 80 = 9.8X t^2. hence 80/9.8 = t^2.hence t= 2.80 5 sec. Now contained in the horizontal body: initial speed no longer regularly occurring. the speed consistently continues to be consistent because the acceleration is contained in the vertical body and under no circumstances contained in the horizontal body. you'd be wanting to keep the vertical and horizontal frames separate and manage them one after the other and your existence will change into undemanding. Time consistently continues to be the same in both the frames. Time X consistent horizontal speed = distance. Distance is 50 meters hence 50= V X time = V X 2.80 5. hence V or the horizontal consistent speed is= 50/2.80 5= 17.fifty 4 meters/sec. i did not choose that 40 5 degrees./ it would want to no longer be proper both!