A ball is thrown horizontally at a speed of 5.8 m/s from the top of a wall 154.4 m high. time for ball to land?
distance from base of wall?
vertical component of velociry just before impact?
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answers:
oubaas say: A ball is thrown horizontally at a speed of 5.8 m/s from the top of a wall 154.4 m high.
time t for ball to land?
h = Voy*t+g/2*t^2
since the ball is thrown horizontally ,then Voy = 0 and the equation above becomes :
h = g/2*t^2
which leads to :
t = √2h/g = √154.4*2/9.806 = 5.61 sec
distance d from base of wall?
d = Vox*t = 5.8*5.61 = 32.5 m
vertical component of velocity Vy just before impact?
Vy = g*t = 9.806*5.61 = 55.0 m/sec
velocity V just before impact?
V = √Vox^2+2gh = √5.8^2+308.8*9.806 = 55.3 m/sec
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electron1 say: Let’s use the following equation to determine the time.
d = vi * t + ½ * g * t^2
d = 154.4 meters
vi = initial vertical velocity
Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. a is 9.8 m/s^2.
154.4 = ½ * 9.8 * t^2
t = √(154.4 ÷ 4.9)
This is approximately 5.6 seconds. Let’s use time in the initial horizontal velocity to determine distance from base of wall.
d = v * t
d = 5.8 * √(154.4 ÷ 4.9)
This is approximately 33 meters. Let’s use the following equation to determine the ball’s vertical velocity just before it hits the ground.
vf = vi + g * t, vi = 0
vf = 9.8 * √(154.4 ÷ 4.9)
This is approximately 55 m/s. I hope this is helpful for you. I rounded all of the answers to two significant digits.
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Andrew Smith say: A standard problem. The horizontal velocity does not affect the vertical motion.
Vertically s = ut + 1/2 a t^2
but the initial vertical velocity is zero
so h = 1/2 g t^2
t = sqrt( 2h/g) ~= sqrt( 2 *154.4 / 9.8) [ 5.6 s ]
Now we know the time and the horizontal distance = vt
= 5.8 * t [32.6 m]
The vertical component of the velocity is gt [ 55 m/s]
The magnitude of the velocity ie speed at landing =sqrt ((gt)^2 + u^2)
= sqrt( 55^2 + 5.8^2) [55.3 m/s]
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gxtzr say: vshgjcjb
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Berger say: nenkqcem
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