Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?
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distance traveled.
Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?
The range or horizontal displacement of a projectile depends on both the horizontal and vertical components of velocity.
While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers away from the original site. Assuming that they were at a height of 137 meters, calculate the horizontal velocity of the spacecraft during touchdown if it lands in a free-fall mode without using retro engines. Consider gravity = 1.63 meters/second2.
To calculate the horizontal velocity of the spacecraft during touchdown, we can use the equation:
distance = velocity x time
In this case, we know the distance (4 kilometers = 4000 meters) and the height (137 meters). We can assume that the time of flight is the same for both the horizontal and vertical components of motion, as the spacecraft is in free-fall mode without using retro engines.
First, we can calculate the time of flight using the vertical motion equation:
h = (1/2)gt^2
Where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Plugging in the values, we have:
137 = (1/2)(1.63) t^2
274 = (1.63) t^2
t^2 = 274 / 1.63
t^2 = 168.22
t ≈ √168.22
t ≈ 12.98 seconds
Now, let's calculate the horizontal velocity using the equation:
distance = velocity x time
4000 = velocity * 12.98
velocity = 4000 / 12.98
velocity ≈ 308.32 meters/second
Therefore, the horizontal velocity of the spacecraft during touchdown is approximately 308.32 meters/second.
distance = velocity x time
In this case, we know the distance (4 kilometers = 4000 meters) and the height (137 meters). We can assume that the time of flight is the same for both the horizontal and vertical components of motion, as the spacecraft is in free-fall mode without using retro engines.
First, we can calculate the time of flight using the vertical motion equation:
h = (1/2)gt^2
Where h is the height, g is the acceleration due to gravity, and t is the time of flight.
Plugging in the values, we have:
137 = (1/2)(1.63) t^2
274 = (1.63) t^2
t^2 = 274 / 1.63
t^2 = 168.22
t ≈ √168.22
t ≈ 12.98 seconds
Now, let's calculate the horizontal velocity using the equation:
distance = velocity x time
4000 = velocity * 12.98
velocity = 4000 / 12.98
velocity ≈ 308.32 meters/second
Therefore, the horizontal velocity of the spacecraft during touchdown is approximately 308.32 meters/second.
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