Respuesta :
1) The law of motion of the projectile is
[tex]h(t) = 2+22.5 t-4.9 t^2[/tex]
To find the velocity, we should compute the derivative of h(t):
[tex]v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t[/tex]
So now we can calculate the speed at t=2 s and t=4 s:
[tex]v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s[/tex]
[tex]v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s[/tex]
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
[tex]v(t)=0[/tex]
So we have
[tex]22.5-9.8 t=0[/tex]
And solving this we find
[tex]t=2.30 s[/tex]
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
[tex]h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m[/tex]
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
[tex]2+22.5t -4.9 t^2 = 0[/tex]
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
[tex]t=4.68 s[/tex]
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
[tex]v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s[/tex]
with negative sign, because it is directed downward.
[tex]h(t) = 2+22.5 t-4.9 t^2[/tex]
To find the velocity, we should compute the derivative of h(t):
[tex]v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t[/tex]
So now we can calculate the speed at t=2 s and t=4 s:
[tex]v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s[/tex]
[tex]v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s[/tex]
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
[tex]v(t)=0[/tex]
So we have
[tex]22.5-9.8 t=0[/tex]
And solving this we find
[tex]t=2.30 s[/tex]
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
[tex]h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m[/tex]
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
[tex]2+22.5t -4.9 t^2 = 0[/tex]
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
[tex]t=4.68 s[/tex]
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
[tex]v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s[/tex]
with negative sign, because it is directed downward.
The velocity of the object when it hits the ground is 23.36 m/s.
When object hits the ground it's height [tex]\rm \bold { H_(_t_) = 0}[/tex], as given here
[tex]\rm \bold{ h= 2 + 22.5t- 4.9t^2 }\\\\\rm \bold{ 0= 2 + 22.5t- 4.9t^2 }\\\\[/tex]
we get 2 values for this equation. first is negative hence, ignored.
t = 4.68 s
So, the object hits the ground at 4.68 s.
The formula for velocity of falling object,
[tex]\rm \bold{ V =V_0- gt}[/tex]
Where,
[tex]\rm \bold{V_0}[/tex] -initial velocity = 22.5 m/s
g - gravitational acceleration = [tex]\rm \bold{ 9.81 m/s^2}[/tex]
t - time = 4.68 s
Put the value in the formula,
[tex]\rm \bold{ V = 22.5 - 9.81m/s^2\times 4.68 s}\\\\\rm \bold{ V = -23.36 m/s}[/tex]
Hence, we can conclude that the velocity of the object when it hits the ground is 23.36 m/s.
To know more about velocity, refer to the link:
https://brainly.com/question/13639113?referrer=searchResults
