Question

An object is thrown upward from the edge of a building with a velocity of 20 m/s. Where will the object be 3s after it was thrown?

−22m
16m
22m

−16m

Answer 1

Answer on Question #51859, Physics, Mechanics | Kinematics | Dynamics

Given:

$$v _ {0} = 2 0 m / s \quad t = 3 s \quad \alpha = 9 0 ^ {0}$$

Find: $h$

Solution:

Lets $t_{AC} = t = 3s$

$$t _ {A D} = 2 \frac {v _ {0} \sin \alpha}{g} = 4. 0 8 s$$ $$\mathrm {h} _ {\mathrm {A B}} = \frac {v _ {0} ^ {2} \sin^ {2} \alpha}{2 g} = 2 0. 4 m$$ $$t _ {B D} = \sqrt {\frac {2 h _ {A B}}{g}} = 2. 0 2 s$$ $$t _ {A B} = t _ {A D} - t _ {B D} = 2. 0 6 s$$ $$t _ {B C} = t _ {A C} - t _ {A B} = 0. 9 4 s$$ $$\mathrm {h} _ {\mathrm {B C}} = \frac {g t _ {B C} ^ {2}}{2} = \frac {9 . 8 \cdot 0 . 8 8 3 6}{2} m = 4. 3 3 m$$ $$h = h _ {A B} - h _ {B C} = 2 0. 4 m - 4. 3 3 m = 1 6. 0 7 m \approx 1 6 m$$

Answer: $h = 16m$

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