A projectile is launched horizontally from a height of 2.2 meters at 35 m/s. How far will it travel horizontally before it hits the ground?
how long will it take to hit the ground?
2.2 = (1/2) (9.81) t^2
solve for t
then
d = u t = 35 t
To find the horizontal distance traveled by the projectile before it hits the ground, we can use the equation of motion:
\[d = v_x \cdot t\]
Where:
- \(d\) is the horizontal distance traveled (what we want to find)
- \(v_x\) is the horizontal component of the velocity (which is the initial velocity of the projectile)
- \(t\) is the time it takes for the projectile to reach the ground
To calculate the time it takes for the projectile to reach the ground, we can use the formula:
\[t = \sqrt{\frac{2h}{g}}\]
Where:
- \(h\) is the initial height of the projectile
- \(g\) is the acceleration due to gravity, which is approximately \(9.8 \, \text{m/s}^2\)
Plugging in the values given in the problem:
\(h = 2.2 \, \text{m}\)
\(v_x = 35 \, \text{m/s}\)
\(g = 9.8 \, \text{m/s}^2\)
We can calculate the time:
\[t = \sqrt{\frac{2 \times 2.2}{9.8}}\]
Simplifying:
\[t = \sqrt{\frac{4.4}{9.8}}\]
Taking the square root:
\[t \approx 0.688 \, \text{s}\]
Now we can use the equation of motion to find the horizontal distance traveled:
\[d = v_x \cdot t\]
Plugging in the values:
\[d = 35 \, \text{m/s} \times 0.688 \, \text{s}\]
Calculating:
\[d \approx 24.08 \, \text{m}\]
Therefore, the projectile will travel approximately 24.08 meters horizontally before hitting the ground.