Question

neglecting air resistance, if a ball thrown 4.5 m/s horizontally from a 94 m cliff, how far has the ball fallen after 2.7 seconds?

Answer 1

Neglecting air resistance, if a ball thrown $4.5\mathrm{m / s}$ horizontally from a $94\mathrm{m}$ cliff, how far has the ball fallen after 2.7 seconds?

Solution:

Equation of motion of the ball along the X-axis:

$S = V_{x}t + \frac{g_{x}t^{2}}{2}, g_{x} - \text{the projection of the gravitational acceleration on the X axis}$

$$g _ {x} = 0, V _ {x} = V, S = V t = 4. 5 \frac {m}{s} * 2. 7 s = 1 2. 1 5 m$$

Equations of motion along the Y-axis:

$h = V_{y}t + \frac{g_{y}t^{2}}{2}, g_{y} - \text{the projection of the gravitational acceleration on the Y axis}$

$$g _ {y} = g, V _ {y} = 0, h = \frac {g t ^ {2}}{2} = \frac {9 . 8 \frac {m}{s ^ {2}} * 2 . 7 s * 2 . 7 s}{2} = 3 5. 7 2 1 m$$

The distance from the point of throwing by the Pythagorean theorem:

$$L = \sqrt {h ^ {2} + S ^ {2}} = 3 7. 7 3 m$$

Distance above the ground: $k = H - h = 94 \, m - 35.721 \, m = 58.279 \, m$

Answer: distance above the ground: $k = 58.279 \, \text{m}$ ;

distance from the point of throwing: $L = 37.73 \, \text{m}$ ;

height of ball fall: $h = 35.721 \, \text{m}$ ;

length of the path of the ball along the X axis: $S = 12.15 \, \text{m}$ .