Bob has just finished climbing a sheer cliff above a level beach and wants to figure out how high he climbed. All he has to use is baseball, a stopwatch, and friend on the ground below with a long measuring tape. Bob is a pitcher and knows that the fastest he can throw the ball is about v_0= 34 m/s. Bob starts the stopwatch as he throws the ball, with no way to measure the ball’s initial trajectory, and he watches carefully. The ball rises and then falls, and after t_1= 0.510 s the ball is once again level with Bob. Bob cannot see we’ll enough to time when the ball hits the ground. Bob’s friend the measures that the ball hit the ground x= 127 m from the base of the cliff. How high above the beach was the ball when it was thrown? A projectile is launched from ground level with an initial speed of 45.5 m/s at an angle of 32.3 degrees above the horizontal. It strikes a target in the air 1.83 s later. What is the horizontal distance from where the projectile was launched to where it hits the target? What is the vertical distance from where the projectile was launched to where it hits the target?
The correct answer and explanation is :
Part 1: Bob and the Ball
Given:
- Maximum speed Bob can throw: ( v_0 = 34 \, \text{m/s} )
- Time until the ball returns to Bob’s level: ( t_1 = 0.510 \, \text{s} )
- Horizontal distance from cliff base to where the ball lands: ( x = 127 \, \text{m} )
- We are asked to find the height of the cliff, i.e., the vertical distance from where the ball was thrown to the ground.
Step 1: Find Vertical Velocity Component
Since the ball returns to Bob’s level in ( t_1 = 0.510 \, \text{s} ), this is the time to the top and back down to his height — total time at Bob’s level. So the upward time is half, i.e. ( t_{\text{up}} = 0.255 \, \text{s} ).
Use the vertical motion formula:
[
v_{0y} = g \cdot t_{\text{up}} = 9.8 \cdot 0.255 = 2.499 \, \text{m/s}
]
So vertical velocity: ( v_{0y} \approx 2.50 \, \text{m/s} )
Step 2: Find Horizontal Velocity Component
Using Pythagoras on the velocity components:
[
v_{0x} = \sqrt{v_0^2 – v_{0y}^2} = \sqrt{34^2 – 2.5^2} \approx 33.91 \, \text{m/s}
]
Step 3: Total Flight Time
Use the horizontal motion to find total time:
[
t = \frac{x}{v_{0x}} = \frac{127}{33.91} \approx 3.745 \, \text{s}
]
Step 4: Height of Cliff
Use vertical displacement formula from launch to ground:
[
y = v_{0y} \cdot t – \frac{1}{2} g t^2 = 2.5 \cdot 3.745 – \frac{1}{2} \cdot 9.8 \cdot (3.745)^2
]
[
y \approx 9.36 – 68.86 = -59.5 \, \text{m}
]
So, height of the cliff is approximately 59.5 m.
Part 2: Projectile at 32.3°
Given:
- ( v_0 = 45.5 \, \text{m/s} )
- Angle ( \theta = 32.3^\circ )
- Time ( t = 1.83 \, \text{s} )
Step 1: Horizontal and Vertical Velocity Components
[
v_{0x} = v_0 \cos\theta = 45.5 \cos(32.3^\circ) \approx 38.5 \, \text{m/s}
]
[
v_{0y} = v_0 \sin\theta = 45.5 \sin(32.3^\circ) \approx 24.3 \, \text{m/s}
]
Step 2: Horizontal Distance
[
x = v_{0x} \cdot t = 38.5 \cdot 1.83 \approx 70.5 \, \text{m}
]
Step 3: Vertical Distance
[
y = v_{0y} \cdot t – \frac{1}{2} g t^2 = 24.3 \cdot 1.83 – 4.9 \cdot 1.83^2
]
[
y \approx 44.5 – 16.4 = 28.1 \, \text{m}
]
Final Answers:
- Cliff Height: 59.5 m
- Projectile Horizontal Distance: 70.5 m
- Projectile Vertical Distance: 28.1 m
Explanation (300 words)
In both scenarios, we apply fundamental kinematic equations that describe motion in two dimensions. For Bob’s problem, the key insight is that the total time the ball takes to reach the ground can be found using horizontal motion, since vertical motion is complex due to the unknown height. Bob throws the ball horizontally and sees it return after 0.510 seconds. That tells us the vertical velocity component based on gravity and time. From that and the total speed, we deduce the horizontal component. Since we know the horizontal distance and speed, we compute the total time it took for the ball to hit the ground, and then calculate the vertical distance it fell during that time using vertical motion equations. The result is the height of the cliff.
In the projectile problem, we break the motion into horizontal and vertical components using trigonometry. The horizontal component of velocity remains constant (ignoring air resistance), and so we can easily calculate how far it travels in the given time. Vertically, the object follows a parabolic path, first rising and then falling under gravity. By calculating the vertical displacement, we find how far above the launch point the projectile is when it hits the target. This decomposition method is critical in all projectile motion problems and is rooted in the independence of vertical and horizontal motions.