A projectile is fired horizontally from a gun that is 96.0 m above flat ground,?

Question

emerging from the gun with a speed of 250 m/s.
a.At what horizontal distance from the firing point does it strike the ground?
m
b.What is the magnitude of the vertical component of its velocity as it strikes the ground?
m/s
Answer:
1107m for a
43.4m/s for b
just show the work

Answer 1

x= -g/2t²+v0t+x0
0= -9.8/2t²+96 (we are only calculating the freefall time to hit the ground. Horizontal speed is irrelevant)
t²=19.59
t=4.426266681

distance traveled in 4.4... sec = 250*4.4...=1107
(Sorry about not putting all those decimal places in there, but I'm lazy)

The vertical speed is -9.8m/sec*4.426266681 = 43.4

Note: the muzzle velocity determined the distance traveled, but not the time to hit the ground nor the vertical component of speed.

Answer 2

a. height above ground = s = 96 m

First find the time it takes for the projectile to reach the ground which is the time of flight.
s = (1/2)g t^2 where g = acceleration of gravity = 9.8 m/s^2
t = SQRT(2s/g)
t = SQRT(2*96/9.8) = SQRT(19.5918)
t = 4.4263 s

Horizontal distance = (horizontal speed)x(time of flight)
Distance = 250(4.4263)
Distance = 1106.5667 meters

b.v = v0 + gt
v0 = 0 since the projectile starts out only moving horizontally and then being pulled down by gravity

so: v = gt = 9.8*4.4263
v = 43.3774 m/s

Answer 3

In the vertical direction :
u = 0 m/s (since the initial vertical velocity is 0)
a = 9.8
s = 96
Use s = 1/2 * at^2
t = 4.43 s
v = u + at = at = 43.4 m/s

In the horizontal direction :
t = 4.43 s
v = 250 m/s
s = vt = 1107 m

Hope this helps.

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Answer 4

Take downwqard direction as advantageous: question a) this is going to proceed to be interior the air for the time it takes to fall 55m vertically downwards: s = ut + ½*a*t² fifty 5 = 0 + ½*9.8*t² 4.9t² = fifty 5 t² = fifty 5/4.9 t² = 11.22 t = 3.35 seconds Projectile is interior the air for 3.35 seconds question b) Distance = velocity *time Distance = 230*3.35 Distance = 770m horizontally question c) v = u +at v = 0 +9.8*3.35 v = 32.8m/s

Answer 5

vertical motion:
u=0,a=+10m/s^2,s=96m,t=?, v=?
a) using S=ut +1/2 a t^2
96=0+5t^2
t^2=19.2
t=4.382s
b) using v=u +at,
v=0+10*4.382
=43.8 2m/s
horizontal distance covered=250*4.382
= 1095.45m
better use a=9.8m/s^2 instead of 10m/s^2