A bullet is fired in a horizontal direction with a muzzle velocity of 300m\s in the absence of air resistance,how far will it have dropped in travelling a horizontal distance of (a)20m (b)40m (c)60m (d)how far will it drop in one second
a. Dx = Xo*t. t = Dx/Xo = 20/300 = 0.067 s.
Dy = 0.5g*t^2 = 4.9*0.067^2 = 0.022 m.
b. t = d/Vo = 40/300 = 0.133 s.
Dy = 4.9*0.133^2 = 0.087 m.
c. t = Dx/Vo = 80/300 = 0.267 s.
Dy = 4.9*0.267^2 = 0.348 m.
d. Dy = 4.9*1^2 = 4.9 m.
(a) S = ut+1/2at² (a=0)
S=ut
t= s/u
t= 20/300
t= 0.067 sec
h= ut+1/2gt² ( u=0)
h= 1/2*5*(0.067)²
h= 0.022 m
(b)s should be equal to 40m
(c) S = 60m
(d) h=ut+1/2gt² (u=0)
h=1/2gt² (g=9.8 or 10)
h= 1/2*9.8*1
h= 4.9m ( if g=10, h=5m)
I don't understand
Please explain
Them
Number b and c u did not explain them
How do you get 4.9
20×3=
To determine how far the bullet will have dropped, let's break down the problem using basic physics principles.
Given:
- Muzzle velocity of the bullet (horizontal direction): 300 m/s
- No air resistance
Now, let's solve it step by step:
Step 1: Determine the time of flight:
Since there is no air resistance, the horizontal velocity remains constant throughout the trajectory. Therefore, the time taken to cover the given horizontal distance will be the same for all cases.
Time = Distance / Velocity
(a) Time taken to cover 20m: 20m / 300 m/s = 1/15 s
(b) Time taken to cover 40m: 40m / 300 m/s = 2/15 s
(c) Time taken to cover 60m: 60m / 300 m/s = 1/5 s
Step 2: Determine the vertical distance dropped:
The bullet will experience free fall vertically due to gravity while traveling horizontally. The distance it drops can be calculated using the formula:
Vertical distance dropped (d) = (1/2) * g * t^2
where g is the acceleration due to gravity (usually approximated as 9.8 m/s^2) and t is the time of flight.
(a) Vertical distance dropped for 20m: d = (1/2) * 9.8 m/s^2 * (1/15 s)^2
(b) Vertical distance dropped for 40m: d = (1/2) * 9.8 m/s^2 * (2/15 s)^2
(c) Vertical distance dropped for 60m: d = (1/2) * 9.8 m/s^2 * (1/5 s)^2
Step 3: Calculate the result:
(a) Vertical distance dropped for 20m = (1/2) * 9.8 m/s^2 * (1/15 s)^2 = 0.01307 m
(b) Vertical distance dropped for 40m = (1/2) * 9.8 m/s^2 * (2/15 s)^2 = 0.0523 m
(c) Vertical distance dropped for 60m = (1/2) * 9.8 m/s^2 * (1/5 s)^2 = 0.196 m
To find the distance dropped in one second, we use the same formula but plug in a time of 1 second:
(d) Vertical distance dropped in 1 second: d = (1/2) * 9.8 m/s^2 * (1 s)^2 = 4.9 m
So, the answers to your questions are:
(a) The bullet will drop approximately 0.01307 meters while traveling 20 meters horizontally.
(b) The bullet will drop approximately 0.0523 meters while traveling 40 meters horizontally.
(c) The bullet will drop approximately 0.196 meters while traveling 60 meters horizontally.
(d) The bullet will drop approximately 4.9 meters in one second.