MICROWAVE
1 answer:
You might be interested in
(1) The initial velocity of the ball is 7.8 m/s.
(2a) The time of motion of the baseball is 6.5 seconds.
(2b) The horizontal distance traveled by the baseball is 206.8 m.
(3) The height of fall of the bullet before hitting the target is 0.078 m.
<h3>Initial velocity of the ball</h3>
The given parameters include;
- height reached by the ball, h = 8 m
- horizontal distance of the ball, x = 10 m
h = Vyt + ¹/₂gt²
where;
- Vy is initial vertical velocity = 0
h = 0 + ¹/₂gt²
t = √(2h/g)
t = √(2 x 8)/9.8
t = 1.28 s
x = Vxt
Vx = x/t
Vx = 10/1.28
Vx = 7.8 m/s
<h3>Time of motion of the base ball</h3>The given parameters include;
- initial velocity, u = 45 m/s
- angle of projection, θ = 45⁰
x = Vxt
x = (ucosθ)t
x = (45 cos(45)) x (6.5)
x = 206.8 m
<h3>Height reached by the bullet</h3>The given parameters include;
- horizontal velocity of the bullet, Vx = 800 m/s
- Horizontal distance of the bullet, x = 100 m
x = Vxt
t = x/Vx
t = 100/800
t = 0.125 s
h = ¹/₂gt²
h = ¹/₂(9.8)(0.125)²
h = 0.078 m
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1