Superman is said to be able to ''leap buildings in a single bound'' How high a building could Superman jump over if he were to leave the ground with speed of 98.0 m/s at an angle of 87.0 degrees to the horizontal?
h = -1/2 g t^2 + [98.0 sin(87.0º)]
the max is on the axis of symmetry ... t = -b / 2a = -[98.0 sin(87.0º)] / -g
solve for t ... plug into h equation to find max height
To determine the height of the building that Superman could jump over, we can use basic principles of projectile motion. Let's break down the problem into two components: vertical motion and horizontal motion.
1. Vertical Motion:
When Superman jumps, the vertical component of his initial velocity determines how high he can go. We can use the following equation to find the maximum height (H) reached by an object in projectile motion:
H = (v₀² * sin²θ) / (2 * g)
Where:
- H is the maximum height
- v₀ is the initial velocity
- θ is the launch angle
- g is the acceleration due to gravity (approximately 9.8 m/s²)
Plugging in the given values, we have:
v₀ = 98.0 m/s
θ = 87.0 degrees
g = 9.8 m/s²
Convert the angle from degrees to radians:
θ = 87.0 degrees * (π/180 radians/degree) ≈ 1.51844 radians
Calculate the maximum height (H):
H = (98.0² * sin²1.51844) / (2 * 9.8)
2. Horizontal Motion:
The horizontal component of Superman's velocity determines the distance he can travel. We can calculate the horizontal distance (D) using the following equation:
D = (v₀ * sin(2θ)) / g
Plugging in the given values:
v₀ = 98.0 m/s
θ = 87.0 degrees
g = 9.8 m/s²
Convert the angle from degrees to radians:
θ = 87.0 degrees * (π/180 radians/degree) ≈ 1.51844 radians
Calculate the horizontal distance (D):
D = (98.0 * sin(2 * 1.51844)) / 9.8
By solving these equations, we can determine both the maximum height and horizontal distance Superman can achieve.