Physics, 17.07.2019 00:00 janaih

Abaseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will the ball hit the ground? (neglect air friction.) 1.11×101 m you are correct. your receipt no. is 153-2897 previous tries what is the speed of the baseball when it hits the ground?

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1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t

$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s

, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2

, and initial height h=1.85 m

.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0

:
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
v_x=18.1 m/s

On the y-axis, the velocity is given by
v_y(t)=gt

If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s

And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

Answer from: samantha636

11.117m; 19.08m/s

Explanation:

Range is the distance covered by the body in horizontal direction.

Range = u × √2H/g where:

u is the initial speed of the base ball = 18.1m/s

H is the maximum height reached = 1.85m

g is the acceleration due to gravity = 9.81m/s²

R = 18.1 × √2(1.85)/9.81

R = 18.1 × √3.70/9.81

R = 18.1 × √0.3772

R = 18.1 × 0.6142

R = 11.117m

To get the speed of the ball on hitting the ground, we will use the equation of motion

v² = u² + 2gH

v² = 18.1²+2(9.81)(1.85)

v² = 327.61+36.297

v² = 363.907

v = √363.907

v = 19.08m/s

Answer from: Quest

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation:

For which optical devices does d sometimes have a positive value?

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

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Abaseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at wh...

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