At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 53 m horizontally and h = 54 m vertically above the launch point. What are the (a) horizontal and (b)
vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?
u = 53/2 = 26.5 m/s forever
v = Vi - 9.81 t
h = Vi t - 4.9 t^2 = 54 =Vi(2)-4.9(4)
or
54 = 2 Vi - 19.6
so
Vi = 36.8 m/s
==========================
at top, v = 0
0 = 36.8 - 9.81 t
t = 3.75 seconds at top
but we know u = 26.5 m/s
so
D = u t = 26.5 * 3.75
(a) Why did the projectile go to therapy? Because it had some serious height issues! Now, as for the horizontal component of the initial velocity, let's call it Vx. Since there are no horizontal forces acting on the projectile, we can use the equation Vx = d/t and substitute in the values given. So, Vx = 53 m / 2 s = 26.5 m/s.
(b) Ah, the vertical component, let's call it Vy. Now, since the projectile was launched vertically and reached a height of 54 m, we can use the equation h = (1/2)gt^2 + Vy*t, where g is the acceleration due to gravity. As the projectile was launched from ground level, at t = 0 s, h = 0, and when t = 2 s, h = 54 m. Plugging those values in, we get 54 m = (1/2)*g*(2 s)^2 + Vy*(2 s). Let's solve for Vy, shall we? Using some math magic, we find Vy = 9.8 m/s.
(c) Alright, let's calculate the horizontal displacement D at the instant the projectile reaches its maximum height. When the projectile reaches its maximum height, its vertical velocity becomes zero. Using the equation Vf = Vi + gt, where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity, and t is the time taken, we can find the time taken for the projectile to reach its maximum height. Since Vf is zero, we have 0 = Vy + g*t. Solving for t, we find t = -Vy / g.
Now, we can find the horizontal displacement D using the equation D = Vx*t. Substituting the values we calculated earlier, D = (26.5 m/s) * (-9.8 m/s^2) / (-9.8 m/s^2) = -26.5 m. Voila! The horizontal displacement D from the launch point at the instant the projectile reaches its maximum height is -26.5 m.
To find the horizontal and vertical components of the initial velocity, we can use the given information about the displacement at time t = 2.00 s.
(a) The horizontal component of the initial velocity remains constant throughout the projectile's motion. Therefore, the horizontal component of the initial velocity is equal to the displacement divided by the time:
Horizontal component of initial velocity = displacement / time
= 53 m / 2.00 s
= 26.5 m/s
So, the horizontal component of the initial velocity is 26.5 m/s.
(b) The vertical component of the initial velocity changes due to the effect of gravity. We can use the equation for vertical displacement to find the vertical component of the initial velocity:
Vertical displacement = (initial vertical velocity) * time + (0.5 * acceleration due to gravity * time^2)
The initial vertical velocity is what we are trying to find. The acceleration due to gravity can be taken as -9.8 m/s^2 (taking downward direction as negative).
Plugging in the values:
54 m = (initial vertical velocity) * 2.00 s + (0.5 * (-9.8 m/s^2) * (2.00 s)^2
Solving for the initial vertical velocity:
54 m = (initial vertical velocity) * 2.00 s - 19.6 m
(initial vertical velocity) * 2.00 s = 54 m + 19.6 m
(initial vertical velocity) * 2.00 s = 73.6 m
(initial vertical velocity) = 73.6 m / 2.00 s
(initial vertical velocity) = 36.8 m/s
So, the vertical component of the initial velocity is 36.8 m/s.
(c) At the instant it reaches its maximum height, the vertical component of the velocity becomes zero. The horizontal velocity remains the same throughout. Therefore, the time it takes to reach maximum height is half of the total time of flight.
Time to reach maximum height = 2.00 s / 2 = 1.00 s
The horizontal displacement at maximum height can be calculated using the horizontal component of the initial velocity and the time:
Horizontal displacement at maximum height (D) = horizontal component of initial velocity * time to reach maximum height
D = 26.5 m/s * 1.00 s
D = 26.5 m
So, at the instant it reaches its maximum height above ground level, the horizontal displacement from the launch point is 26.5 m.
To find the horizontal and vertical components of the initial velocity of the projectile:
(a) Horizontal component of initial velocity (Vx):
To find the horizontal component of the initial velocity, we can use the formula:
Vx = d / t
where d is the horizontal displacement (53 m) and t is the time taken (2.00 s).
Thus,
Vx = 53 m / 2.00 s = 26.5 m/s
Therefore, the horizontal component of the initial velocity is 26.5 m/s.
(b) Vertical component of initial velocity (Vy):
To find the vertical component of the initial velocity, we need to use the formula for vertical displacement:
h = Vyi * t + (1/2) * g * t^2
where h is the vertical displacement (54 m), Vyi is the initial vertical velocity, t is the time taken (2.00 s), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
We know that at the highest point, the vertical displacement is zero (h = 0). Therefore, we can rearrange the formula to solve for Vyi:
0 = Vyi * t + (1/2) * g * t^2
Simplifying the equation, we have:
(1/2) * g * t^2 = - Vyi * t
Dividing both sides by t, we get:
(1/2) * g * t = - Vyi
Substituting the values, we have:
(1/2) * 9.8 m/s^2 * 2.00 s = - Vyi
Vyi = -9.8 m/s
Since we want the magnitude of the vertical component of initial velocity, we can ignore the negative sign. Therefore, the vertical component of the initial velocity is 9.8 m/s.
(c) To find the horizontal displacement D when the projectile reaches its maximum height, we can use the formula for horizontal displacement:
D = Vx * t
where Vx is the horizontal component of the initial velocity (26.5 m/s) and t is the time taken to reach maximum height (which can be found using the equation of motion for projectile motion).
To find the time taken to reach maximum height, we know that the vertical component of velocity at maximum height is zero. Using the equation:
Vyf = Vyi + g * t
where Vyf is the final vertical component of velocity, Vyi is the initial vertical component of velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken to reach maximum height.
We can rearrange this equation to solve for t:
0 = Vyi + g * t
Substituting the values, we have:
0 = 9.8 m/s + 9.8 m/s^2 * t
Solving for t, we get:
t = - (9.8 m/s) / (9.8 m/s^2)
t = -1 second
Since time cannot be negative, it means it takes 1 second for the projectile to reach maximum height.
Now, substituting the values into the formula for horizontal displacement:
D = Vx * t
D = 26.5 m/s * 1 s
D = 26.5 m
Therefore, the horizontal displacement D from the launch point at the instant the projectile reaches its maximum height is 26.5 m.