Batman is standing on a cliff 120 m high. He is testing a new bat grappling hook. He

fires the hook with an initial velocity of 85 m/s at an angle of 45o
with respect to the
horizontal. How far from the base of the cliff (d) does the hook land?

Vo = 85m/s[45o].

Xo = 85*cos45 = 60.1 m/s.
Yo = 85*sin45 = 60.1 m/s.

Tr = (Y-Yo)/g = (o-60.1)/-9.8 = 6.1 s.
= Rise time.

h = Yo*t + 0.5g*t^2
h = 60.1*6.1 - 4.9*(6.1)^2 = 184.3 m.

d = Yo*t + 0.5g*t^2=184.3 + 120=304.3m
0 + 4.9t^2 = 304.3
t^2 = 62.1
Tf = 7.9 s. = Fall time.

d =Xo * (Tr + Tf)
d = 60.1*(6.1+7.9) = 841.4 m.


91,75s = 105 m.

To calculate the horizontal distance (d) that the hook lands from the base of the cliff, we can use the kinematic equations of motion.

First, we need to analyze the horizontal and vertical components of the initial velocity. The horizontal component (v_x) can be calculated using the initial velocity (v₀) and the angle (θ):

v_x = v₀ * cos(θ)

In this case, v₀ = 85 m/s and θ = 45°, so:

v_x = 85 m/s * cos(45°)

v_x = 85 m/s * √(2)/2

v_x = 42.5 m/s

Now, we can calculate the time it takes for the hook to hit the ground by considering the vertical motion. The equation for the height (h) as a function of time (t) can be written as:

h = v₀ * sin(θ) * t - (1/2) * g * t²

Where g is the acceleration due to gravity (approximately 9.8 m/s²). We know that the initial height (h) is 120 m and that the final height is 0 m (since the hook lands on the ground).

Setting h = 0, we have:

0 = v₀ * sin(θ) * t - (1/2) * g * t²

Rearranging the equation, we get:

(1/2) * g * t² = v₀ * sin(θ) * t

Simplifying further, we get:

(1/2) * g * t = v₀ * sin(θ)

Now, we can solve for t:

t = 2 * v₀ * sin(θ) / g

Using the values v₀ = 85 m/s, θ = 45°, and g = 9.8 m/s²:

t = 2 * 85 m/s * sin(45°) / 9.8 m/s²

t ≈ 9.244 s

Finally, we can calculate the horizontal distance (d) using the formula:

d = v_x * t

Plugging in the values v_x = 42.5 m/s and t ≈ 9.244 s:

d ≈ 42.5 m/s * 9.244 s

d ≈ 393.53 m

Therefore, the hook lands approximately 393.53 meters away from the base of the cliff.