A stone is thrown horizontally outward from the top of a bridge, the stone is released 19.6meter above the street below.the initial velocity of the stone is 5.0m/s.determine the (a)total time that the stone is in the air

(b)magnitude and direction of the velocity"projectille just before the it strike the street.
(2)A projectile is fired with an initial speed of 113m/s at an angle of 60.0 degree above the horizontal from the top of a cliff 49.0m high.determine(a)the time to reach the maximum height
(b)maximum height above the base of the cliff reached by the projectile
(c) the total time it is in the air
(d)horizontal range of the projectile.plz help

If it is thrown horizontal, the horizontal speed u remains 5 m/s until it stops.

u = 5
v = - g t
h = 19.6 - .5 g t^2
or
0 = 19.6 - 4.9 t^2
t^2 = 19.6/4.9
t = 2 seconds in air

u = 5
v = -9.8 * 2 = -19.6 m/s

so speed at ground = sqrt (25 +19.6^2 )
tan angle below horizontal = 19.6/5

u = 11.3 cos 60 forever

Vi = 11.3 sin 60

v = 11.3 sin 60 - 9.8 t
v = 0 at top so
at top
t = 11.3 sin 60 /9.8
and also at top
h = 49 + Vi t - 4.9 t^2

now at h = 0
0 = 49 + Vi t - 4.9 t^2
solve quadratic for t when h = 0
then that is time in air
range = u t

To solve these problems, we will use the kinematic equations of motion.

Problem 1:
Given:
Initial vertical displacement (y) = -19.6 meters (negative because it is below the reference point)
Initial horizontal velocity (Vx) = 5.0 m/s

(a) Total time the stone is in the air:
Since the stone is thrown horizontally, it does not have any vertical acceleration. We can use the equation for vertical displacement to find the time of flight (t):
y = Vyt + (1/2)gt^2

Considering that the initial vertical velocity (Vy) is 0 m/s, and the acceleration due to gravity (g) is -9.8 m/s^2 (negative because it opposes the positive direction), the equation simplifies to:
-19.6 = 0 + (1/2)(-9.8)t^2

Solving for t:
19.6 = 4.9t^2
t^2 = 19.6/4.9
t^2 = 4
t = √4
t = 2 seconds

(b) Magnitude and direction of velocity just before the stone strikes the street:
Since the stone is thrown horizontally, its horizontal velocity remains constant (5.0 m/s). The vertical velocity changes due to gravity. Using the equation for final velocity (Vf = Vi + at), we can find the vertical velocity (Vy) just before striking the street:
Vf = Vi + gt

Considering that Vi = 0 m/s, and t = 2 seconds, the equation simplifies to:
Vf = 0 + (-9.8)(2)
Vf = -19.6 m/s

The magnitude of the velocity just before the stone strikes the street is the resultant of the horizontal and vertical velocities. We can use Pythagoras' theorem to find it:
V = √(Vx^2 + Vy^2)
V = √(5.0^2 + (-19.6)^2)
V = √(25 + 384.16)
V = √409.16
V ≈ 20.23 m/s

The direction of the velocity is given by the angle of the resultant with the horizontal. We can use trigonometry to find the direction:
θ = arctan(Vy / Vx)
θ = arctan(-19.6 / 5.0)
θ ≈ -76.27 degrees (measured from the horizontal)

Therefore, the magnitude of the velocity just before the stone strikes the street is approximately 20.23 m/s. It is directed at an angle of approximately -76.27 degrees (measured from the horizontal).

Problem 2:
Given:
Initial speed (V0) = 113 m/s
Launch angle (θ) = 60.0 degrees
Cliff height (h) = 49.0 m
Acceleration due to gravity (g) = -9.8 m/s^2

(a) Time to reach the maximum height:
The projectile's vertical motion is influenced by gravity. The equation for vertical displacement can be used to find the time to reach the maximum height (tMax):
y = V0yt + (1/2)gt^2

Considering that the initial vertical velocity is Vy0 = V0sinθ, and the final vertical velocity at the maximum height is VyMax = 0 m/s, the equation simplifies to:
h = Vy0tMax + (1/2)gtMax^2

Substituting the given values:
49 = (113 sin 60)tMax + (1/2)(-9.8)tMax^2

Rearranging the equation:
0.5 (-9.8)tMax^2 + (113 sin 60)tMax - 49 = 0

Using the quadratic formula:
tMax = (-b ± √(b^2 - 4ac)) / 2a

where a = 0.5 (-9.8), b = 113 sin 60, and c = -49

Solving for tMax with the positive root:
tMax ≈ 3.06 seconds

(b) Maximum height above the base of the cliff reached by the projectile:
The maximum height can be calculated using the equation for vertical displacement:
yMax = y0 + V0y0tMax + (1/2)gtMax^2

Considering y0 = 0 m and V0y0 = V0 sin θ, we have:
yMax = (113 sin 60)tMax + (1/2)(-9.8)tMax^2
yMax ≈ 90.5 meters

Therefore, the maximum height above the base of the cliff reached by the projectile is approximately 90.5 meters.

(c) Total time the projectile is in the air:
The total time of flight (tTotal) can be found by doubling the time to reach the maximum height:
tTotal = 2tMax
tTotal ≈ 6.12 seconds

(d) Horizontal range of the projectile:
The horizontal range can be calculated using the equation for horizontal displacement:
x = V0x t

Considering V0x = V0 cos θ, we have:
x = (113 cos 60)t
x ≈ 97.7 meters

Therefore, the horizontal range of the projectile is approximately 97.7 meters.

To solve these problems, we can utilize the equations of motion for projectile motion. There are mainly two equations we will use:

1. For vertical motion:
- Final velocity (vf) = Initial velocity (vi) + (acceleration (a) x time (t))
- Displacement (d) = (Initial velocity (vi) x time (t)) + (0.5 x acceleration (a) x (time (t))^2)

2. For horizontal motion:
- Displacement (d) = Initial velocity (vi) x time (t)

Now let's solve each problem step by step.

Problem 1:

a) Total Time:
Since the stone is thrown horizontally, there is no initial vertical velocity. Therefore, there is only vertical acceleration acting on it, which is the acceleration due to gravity (g = 9.8 m/s^2). Using the equation vf = vi + (a x t), we can determine the time.

Initial vertical velocity (vi) = 0 m/s (Thrown horizontally)
Vertical acceleration (a) = -9.8 m/s^2 (Negative because it acts downwards)
Final vertical velocity (vf) = ?

Plug in the values:
vf = vi + (a x t)
vf = 0 + (-9.8 x t)
vf = -9.8t

We know that the final vertical velocity (when the stone hits the ground) is -9.8t. We can set this equation equal to zero to find the time it takes for the stone to hit the ground.

0 = -9.8t
t = 0 seconds or t = any real number.

But we are only interested in the positive time, so the total time the stone is in the air is t = 0 seconds.

b) Magnitude and Direction of Velocity:
Since the stone is thrown horizontally, it will maintain a constant horizontal velocity throughout its flight. The magnitude of the horizontal velocity is given as 5.0 m/s, and the direction is horizontal (parallel to the ground). Therefore, the magnitude and direction of the velocity just before it strikes the street are 5.0 m/s horizontally.

Problem 2:

a) Time to Reach the Maximum Height:
Here, the projectile has an initial vertical velocity, as well as an initial horizontal velocity. We can split the velocities into their horizontal (vx) and vertical (vy) components using trigonometry.

Initial speed (vi) = 113 m/s
Launch angle (θ) = 60.0 degrees

To find the time it takes to reach the maximum height, we need to find the time at which the vertical velocity becomes zero.

Vertical velocity at max height (vy) = 0 m/s
Initial vertical velocity (viy) = vi * sin(θ)
Vertical acceleration (a) = -9.8 m/s^2

Using the equation vf = vi + (a x t), we can calculate the time to reach the maximum height.

vf = viy + (a x t)
0 = viy + (-9.8 x t)
t = -viy / (-9.8)
t = vi * sin(θ) / 9.8

Plug in the values:
t = (113 m/s) * sin(60.0°) / 9.8 m/s^2
t = 6.12 s (approx)

Therefore, the time to reach the maximum height is 6.12 seconds.

b) Maximum Height:
To find the maximum height above the base of the cliff, we need to calculate the vertical displacement using the equation d = (viy x t) + (0.5 x a x t^2).

Initial vertical velocity (viy) = vi * sin(θ)
Time (t) = 6.12 s (from part a)
Vertical displacement (d) = ?

Plug in the values:
d = (vi * sin(θ) x t) + (0.5 x a x t^2)
d = (113 m/s) * sin(60.0°) * 6.12 s + (0.5 x -9.8 m/s^2 x (6.12 s)^2)

Solve the equation to find the displacement.

c) Total Time in the Air:
The total time in the air is given by the time it takes to reach the maximum height, and then doubling that time since the projectile takes the same amount of time to come back down.

Total time = 2 x time to reach the maximum height

time = 2 x 6.12 s

d) Horizontal Range:
The horizontal range is the total horizontal displacement of the projectile. It can be found using the equation d = vix * t, where vix is the horizontal component of the initial velocity (vi * cos(θ)).

Initial horizontal velocity (vix) = vi * cos(θ)
Time (t) = Total time from part c
Horizontal displacement (d) = ?

Plug in the values:
d = (vi * cos(θ)) * (2 x 6.12 s)

Solve the equation to find the horizontal range.

I hope this helps! Let me know if you have any further questions.