Respuesta :
(a) 3.35 s
The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration
a = g = -9.8 m/s^2
towards the ground.
The initial height of the projectile is
h = 55.0 m
The vertical position of the projectile at time t is
[tex]y = h + \frac{1}{2}at^2[/tex]
By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:
[tex]0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s[/tex]
(b) 78.4 m
The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.
The horizontal motion is a uniform motion with constant velocity -
The horizontal velocity of the projectile is
[tex]v_x = 23.4 m/s[/tex]
the time it takes the projectile to reach the ground is
t = 3.35 s
So, the horizontal distance covered by the projectile is
[tex]d=v_x t = (23.4 m/s)(3.35 s)=78.4 m[/tex]
(c) 23.4 m/s, -32.8 m/s
The motion of the projectile consists of two independent motions:
- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to
[tex]v_x = 23.4 m/s[/tex]
so this value is also the value of the horizontal velocity just before the projectile reaches the ground.
- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by
[tex]v_y = u_y +at[/tex]
where
[tex]u_y = 0[/tex] is the initial vertical velocity
Using
a = g = -9.8 m/s^2
and
t = 3.35 s
We find the vertical velocity of the projectile just before reaching the ground
[tex]v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s[/tex]
and the negative sign means it points downward.
The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.
- (a) The time necessary for the projectile to reach the ground below is 3.35 seconds.
- (b) The distance from the base of the building that the projectile lands is 78.4 meters.
- (c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.
What is projectile motion?
Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.
For the motion of horizontal and vertical direction we use following equation in projectile motion.
[tex]y=y_o+\dfrac{1}{2}gt^2[/tex]
Here, (g) is the gravity and (t) is time.
Here, in the given problem, the projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall.
- (a) The time necessary for the projectile to reach the ground below-
As the height of the projectile before the shot is 55 meters and the the height of the projectile is 0 meter when it reaches to the ground. Thus from the above equation,
[tex]0=55+\dfrac{1}{2}\times9.81\times(t)^2\\t=3.35\rm s[/tex]
Thus the time required for the projectile to reach the ground below is 3.35 seconds.
- (b) The distance from the base of the building that the projectile lands-
The distance traveled by the object is the product of velocity and time taken. Thus, distance from the base of the building that the projectile lands is,
[tex]d=23.4\times3.35\\d=78.4\rm m[/tex]
- (c) The horizontal and vertical components of the velocity just before the projectile reaches the ground-
Horizontal component of velocity is equal to the velocity of projectile shot just before it reaches to the ground. Thus, horizontal component of velocity is,
[tex]v_x=23.4\rm m/s[/tex]
Now the time taken is 3.35 seconds. Thus the vertical component of velocity can be found with the formula of first equation of motion with 0 initial velocity as,
[tex]v_y=0+(-9.81)\times(3.35)\\v_y=-32.8\rm m/s[/tex]
The projectile shot from the roof of the building travel in the projectile motion to reach to the ground.
- (a) The time necessary for the projectile to reach the ground below is 3.35 seconds.
- (b) The distance from the base of the building that the projectile lands is 78.4 meters.
- (c) The horizontal and vertical components of the velocity just before the projectile reaches the ground is 23.4 and 32.8 m/s downward respectively.
Learn more about the projectile motion here;
https://brainly.com/question/24216590
