Vv0 =sin(32)*43 m/s
=22.78 m/s
h = Vv0*(t) + 0.5gt^2
if we assume g=10 m/s^2
when the ball lands, h=0
0=22.78*t + 5(t)(t)
t(22.78+5t)=0
t={0s,4.56s}
The ball will be in the air for about 4.5 seconds
To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.
If it's fired horizontally, then its initial vertical velocity is zero. After that, the vertical velocityincreases by 9.8 meters per second every second, directed downward, and the projectile hitsthe ground after roughly 3.8 seconds.Exactly the same vertical motion as if it were dropped from the gun muzzle, with no horizontal velocity.
Hang time depends on your vertical component of velocity when you jump. The higher the vertical velocity, the longer your feet will be off the ground. The horizontal component of velocity does not affect hang time.
The horizontal velocity component of the ball can be found by using the equation: horizontal velocity = initial velocity * cos(angle). In this case, the initial velocity is 26 m/s and the angle is 30 degrees. Plugging in the values, we get: horizontal velocity = 26 m/s * cos(30) ≈ 22.5 m/s.
The thrown ball will have a greater speed when it reaches ground level because it has a horizontal component of velocity in addition to the vertical component. The rock only has a vertical component of velocity due to gravity.
To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
If it's fired horizontally, then its initial vertical velocity is zero. After that, the vertical velocityincreases by 9.8 meters per second every second, directed downward, and the projectile hitsthe ground after roughly 3.8 seconds.Exactly the same vertical motion as if it were dropped from the gun muzzle, with no horizontal velocity.
Hang time depends on your vertical component of velocity when you jump. The higher the vertical velocity, the longer your feet will be off the ground. The horizontal component of velocity does not affect hang time.
The horizontal velocity component of the ball can be found by using the equation: horizontal velocity = initial velocity * cos(angle). In this case, the initial velocity is 26 m/s and the angle is 30 degrees. Plugging in the values, we get: horizontal velocity = 26 m/s * cos(30) ≈ 22.5 m/s.
The thrown ball will have a greater speed when it reaches ground level because it has a horizontal component of velocity in addition to the vertical component. The rock only has a vertical component of velocity due to gravity.
Using the projectile motion equations and given the initial velocity and angle, we can calculate the time the shell is in the air. Then, we can find the horizontal range by multiplying the time of flight by the horizontal component of the initial velocity. The horizontal range in this case is about 1056 meters.
If you're willing to ignore the effect of air resistance, then the answer is as follows: The object's horizontal velocity remains constant (at least until it eventually hits the ground). The vertical component of the object's initial velocity ... call it V(i) ... is the (total initial velocity) multipled by the (sine of the initial angle above the horizontal). Beginning at the time of the toss, the magnitude of the vertical component of velocity is V = V(i) - 1/2gT2. T = number of seconds after the toss g = acceleration of gravity = approx 32 ft/sec2 or 9.8 m/sec2
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
The initial velocity of a projectile affects its range by determining how far the projectile will travel horizontally before hitting the ground. A higher initial velocity will result in a longer range because the projectile has more speed to overcome air resistance and travel further. Conversely, a lower initial velocity will result in a shorter range as the projectile doesn't travel as far before hitting the ground.
The object's initial distance above the ground The object's initial velocity