QuestionA cannon is fired horizontally from a height of $5.0 \mathrm{~m}$ . How long until it hits the ground?

Studdy Solution

STEP 1

Assumptions1. The cannonball is fired horizontally, so its initial vertical velocity is0. . The height of the tower is5.0 m.
3. The acceleration due to gravity is 9.8 \, m/s^.
4. We are neglecting air resistance.

STEP 2

We will use the equation of motion to solve this problem. The equation that relates distance, initial velocity, time, and acceleration is $d = v_i t + \frac{1}{2} a t^2$

STEP 3

In this case, the initial vertical velocity ( $v_i$ ) is0, and the distance (d) is the height of the tower. So the equation simplifies to $d = \frac{1}{2} a t^2$

STEP 4

We can solve this equation for time (t) by multiplying both sides by2, then dividing by acceleration (a), and finally taking the square root of both sides $t = \sqrt{\frac{2d}{a}}$

STEP 5

Now, we can plug in the given values for the distance (d =5.0 m) and acceleration (a =9.8 m/s²) to calculate the time $t = \sqrt{\frac{2 \times5.0 \, m}{9.8 \, m/s^2}}$

STEP 6

Calculate the time it takes for the cannonball to hit the ground $t = \sqrt{\frac{10 \, m}{9.8 \, m/s^2}} \approx1.01 \, s$ The cannonball will hit the ground after approximately1.01 seconds.

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QuestionA cannon is fired horizontally from a height of $5.0 \mathrm{~m}$ . How long until it hits the ground?

Studdy Solution

STEP 1

Assumptions1. The cannonball is fired horizontally, so its initial vertical velocity is0. . The height of the tower is5.0 m.
3. The acceleration due to gravity is 9.8 \, m/s^.
4. We are neglecting air resistance.

STEP 2

We will use the equation of motion to solve this problem. The equation that relates distance, initial velocity, time, and acceleration is $d = v_i t + \frac{1}{2} a t^2$

STEP 3

In this case, the initial vertical velocity ( $v_i$ ) is0, and the distance (d) is the height of the tower. So the equation simplifies to $d = \frac{1}{2} a t^2$

STEP 4

We can solve this equation for time (t) by multiplying both sides by2, then dividing by acceleration (a), and finally taking the square root of both sides $t = \sqrt{\frac{2d}{a}}$

STEP 5

Now, we can plug in the given values for the distance (d =5.0 m) and acceleration (a =9.8 m/s²) to calculate the time $t = \sqrt{\frac{2 \times5.0 \, m}{9.8 \, m/s^2}}$

STEP 6

Calculate the time it takes for the cannonball to hit the ground $t = \sqrt{\frac{10 \, m}{9.8 \, m/s^2}} \approx1.01 \, s$ The cannonball will hit the ground after approximately1.01 seconds.

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QuestionA cannon is fired horizontally from a height of 5.0 m5.0 \mathrm{~m}5.0 m. How long until it hits the ground?

Studdy Solution

STEP 1

Assumptions1. The cannonball is fired horizontally, so its initial vertical velocity is0. . The height of the tower is5.0 m.3. The acceleration due to gravity is 9.8 \, m/s^.4. We are neglecting air resistance.

STEP 2

We will use the equation of motion to solve this problem. The equation that relates distance, initial velocity, time, and acceleration isd=vit+12at2d = v_i t + \frac{1}{2} a t^2d=vi​t+21​at2

STEP 3

In this case, the initial vertical velocity (viv_ivi​) is0, and the distance (d) is the height of the tower. So the equation simplifies tod=12at2d = \frac{1}{2} a t^2d=21​at2

STEP 4

We can solve this equation for time (t) by multiplying both sides by2, then dividing by acceleration (a), and finally taking the square root of both sidest=2dat = \sqrt{\frac{2d}{a}}t=a2d​​

STEP 5

Now, we can plug in the given values for the distance (d =5.0 m) and acceleration (a =9.8 m/s²) to calculate the timet=2×5.0 m9.8 m/s2t = \sqrt{\frac{2 \times5.0 \, m}{9.8 \, m/s^2}}t=9.8m/s22×5.0m​​

STEP 6

Calculate the time it takes for the cannonball to hit the groundt=10 m9.8 m/s2≈1.01 st = \sqrt{\frac{10 \, m}{9.8 \, m/s^2}} \approx1.01 \, st=9.8m/s210m​​≈1.01sThe cannonball will hit the ground after approximately1.01 seconds.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionA cannon is fired horizontally from a height of $5.0 \mathrm{~m}$ . How long until it hits the ground?

Assumptions1. The cannonball is fired horizontally, so its initial vertical velocity is0. . The height of the tower is5.0 m.
3. The acceleration due to gravity is 9.8 \, m/s^.
4. We are neglecting air resistance.

We will use the equation of motion to solve this problem. The equation that relates distance, initial velocity, time, and acceleration is $d = v_i t + \frac{1}{2} a t^2$

In this case, the initial vertical velocity ( $v_i$ ) is0, and the distance (d) is the height of the tower. So the equation simplifies to $d = \frac{1}{2} a t^2$

We can solve this equation for time (t) by multiplying both sides by2, then dividing by acceleration (a), and finally taking the square root of both sides $t = \sqrt{\frac{2d}{a}}$

Now, we can plug in the given values for the distance (d =5.0 m) and acceleration (a =9.8 m/s²) to calculate the time $t = \sqrt{\frac{2 \times5.0 \, m}{9.8 \, m/s^2}}$

Calculate the time it takes for the cannonball to hit the ground $t = \sqrt{\frac{10 \, m}{9.8 \, m/s^2}} \approx1.01 \, s$ The cannonball will hit the ground after approximately1.01 seconds.