- A ball is dropped off the roof of a 50 meter tall building
a. How long does it take to hit the ground?
b. What is the velocity at the moment just before it hits the ground? - If a person steps off a high diving board and falls for 1.12 seconds before he hits the water, how high was the dive?
- A ball is launched upward at a speed of 40 m/s
a. Calculate the maximum height that the ball will reach
b. Calculate the total time of flight for the ball from the moment it is launched, until the moment it hits the
ground again - If you dropped a ball off the roof of a building, how long would it take to reach a speed of 30 m/s?
- A projectile is launched horizontally off a 65 meter high cliff with a speed of 25 m/s.
a. Calculate the time it takes to hit the ground
b. Calculate horizontal range (distance) of the projectile
c. What is the speed of the projectile at the moment it hits the ground?
- If a projectile is launched horizontally off a 100 meter high cliff, how fast does it need to be launched in order to hit
a target that is on the ground 120 meters away? - A horizontal projectile launched at 16.5 m/s lands in a cup 12 meters away from the base of the ledge. Calculate the
height of the ledge.
The Correct Answer and Explanation is :
Let’s go through each problem one by one:
1. A ball is dropped off the roof of a 50 meter tall building
a. How long does it take to hit the ground?
We’ll use the kinematic equation for free fall:
[
h = \frac{1}{2} g t^2
]
Where:
- (h = 50) meters (height of the building)
- (g = 9.8 \, \text{m/s}^2) (acceleration due to gravity)
- (t) is the time it takes to hit the ground
Rearranging the equation to solve for (t):
[
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 50}{9.8}} = \sqrt{10.2} \approx 3.19 \, \text{seconds}
]
b. What is the velocity at the moment just before it hits the ground?
We’ll use the equation for velocity in free fall:
[
v = g t
]
Substitute (g = 9.8 \, \text{m/s}^2) and (t = 3.19 \, \text{seconds}):
[
v = 9.8 \times 3.19 \approx 31.3 \, \text{m/s}
]
2. A person steps off a high diving board and falls for 1.12 seconds before he hits the water. How high was the dive?
We use the same free fall equation:
[
h = \frac{1}{2} g t^2
]
Where (g = 9.8 \, \text{m/s}^2) and (t = 1.12) seconds:
[
h = \frac{1}{2} \times 9.8 \times (1.12)^2 \approx 6.13 \, \text{meters}
]
3. A ball is launched upward at a speed of 40 m/s
a. Calculate the maximum height the ball will reach.
At maximum height, the velocity becomes 0. Using the equation:
[
v^2 = u^2 – 2 g h
]
Where:
- (v = 0) m/s (velocity at the top),
- (u = 40) m/s (initial velocity),
- (g = 9.8) m/s².
Rearranging to solve for height (h):
[
h = \frac{u^2}{2g} = \frac{(40)^2}{2 \times 9.8} = \frac{1600}{19.6} \approx 81.63 \, \text{meters}
]
b. Calculate the total time of flight for the ball.
The total time is the time to go up and come down. The time to reach maximum height is:
[
t_{\text{up}} = \frac{u}{g} = \frac{40}{9.8} \approx 4.08 \, \text{seconds}
]
The total time of flight is twice the time to go up:
[
t_{\text{total}} = 2 \times 4.08 \approx 8.16 \, \text{seconds}
]
4. If you dropped a ball off the roof of a building, how long would it take to reach a speed of 30 m/s?
Using the equation for velocity in free fall:
[
v = g t
]
Rearranging for time (t):
[
t = \frac{v}{g} = \frac{30}{9.8} \approx 3.06 \, \text{seconds}
]
5. A projectile is launched horizontally off a 65 meter high cliff with a speed of 25 m/s
a. Calculate the time it takes to hit the ground.
We’ll use the same free fall equation for vertical motion:
[
h = \frac{1}{2} g t^2
]
Solving for (t):
[
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 65}{9.8}} \approx 3.64 \, \text{seconds}
]
b. Calculate the horizontal range (distance) of the projectile.
Since horizontal velocity remains constant, the horizontal range is:
[
\text{Range} = \text{horizontal speed} \times \text{time} = 25 \times 3.64 \approx 91 \, \text{meters}
]
c. What is the speed of the projectile at the moment it hits the ground?
The vertical velocity at impact is:
[
v_{\text{vertical}} = g t = 9.8 \times 3.64 \approx 35.7 \, \text{m/s}
]
The total speed at impact is the vector sum of horizontal and vertical velocities:
[
v_{\text{total}} = \sqrt{v_{\text{horizontal}}^2 + v_{\text{vertical}}^2} = \sqrt{25^2 + 35.7^2} \approx 44.8 \, \text{m/s}
]
6. If a projectile is launched horizontally off a 100-meter high cliff, how fast does it need to be launched in order to hit a target that is on the ground 120 meters away?
First, find the time it takes to hit the ground:
[
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 100}{9.8}} \approx 4.52 \, \text{seconds}
]
Now, find the required horizontal speed:
[
\text{horizontal speed} = \frac{\text{distance}}{\text{time}} = \frac{120}{4.52} \approx 26.5 \, \text{m/s}
]
7. A horizontal projectile launched at 16.5 m/s lands in a cup 12 meters away from the base of the ledge. Calculate the height of the ledge.
We use the time of flight to find the height:
[
t = \frac{\text{distance}}{\text{horizontal speed}} = \frac{12}{16.5} \approx 0.727 \, \text{seconds}
]
Now, use this time to find the height:
[
h = \frac{1}{2} g t^2 = \frac{1}{2} \times 9.8 \times (0.727)^2 \approx 2.7 \, \text{meters}
]
Summary Answers:
- a: 3.19 seconds; b: 31.3 m/s
- 6.13 meters
- a: 81.63 meters; b: 8.16 seconds
- 3.06 seconds
- a: 3.64 seconds; b: 91 meters; c: 44.8 m/s
- 26.5 m/s
- 2.7 meters
Each of these problems involves principles of free fall and projectile motion, where vertical and horizontal motions are treated independently. The acceleration due to gravity affects the vertical motion, while the horizontal motion remains unaffected by gravity (in the absence of air resistance).