Question

A glove is dropped from a 40 m tall building and simultaneously a ball is thrown from the ground at the speed of 40 m/s. when and where do they meet??

Answer 1

Answer on Question#39362 – Physics – Mechanics

A glove is dropped from a 40 m tall building and simultaneously a ball is thrown from the ground at the speed of 40 m/s. when and where do they meet??

Solution:

$V_{1} = 40\frac{\mathrm{m}}{\mathrm{s}}$ — velocity of the stone, which was thrown down;

$V_{2} = 0$ — velocity of the glove, which was thrown up;

$H = 40\mathrm{m}$ — height of the building;

$h$ — height of the point where the stone and glove cross paths (above the base of the building).

$t$ — time after glove and the ball will meet

The equation of motion for the stone (which was thrown up) respect to the Y-axis:

$$y _ {1} = V _ {1} t - \frac {g t ^ {2}}{2}$$

The equation of motion for the glove (which was thrown down) respect to the Y-axis:

$$y _ {2} = H - V _ {2} t - \frac {g t ^ {2}}{2} = H - \frac {g t ^ {2}}{2}$$

When stone and glove cross paths, their coordinates are equal:

$$y _ {2} = y _ {1}$$

(3) and (2) in (1):

$$V _ {1} t _ {\text {c r o s s}} - \frac {g t _ {\text {c r o s s}} ^ {2}}{2} = H - \frac {g t _ {\text {c r o s s}} ^ {2}}{2}$$ $$H = t _ {\text {c r o s s}} V _ {1}$$ $$t _ {\text {c r o s s}} = \frac {H}{V _ {1}} = \frac {4 0 \mathrm {m}}{4 0 \frac {\mathrm {m}}{\mathrm {s}}} = 1 \mathrm {s}$$ $$h = y _ {2} \left(t _ {\text {c r o s s}}\right) = y _ {1} \left(t _ {\text {c r o s s}}\right) = H - \frac {g \left(t _ {\text {c r o s s}}\right) ^ {2}}{2} = 4 0 m - \frac {9 . 8 \frac {m}{s ^ {2}} \cdot 1 s ^ {2}}{2} = 3 5. 1 m$$

The location (above the base of the building) of the point where the paths of the stone and glove will cross:

$$\mathrm {h} = 3 5. 1 \mathrm {m}$$

**Answer**: paths will cross after 1 second at the height $h = 35.1 \, \text{m}$ above the base of the building.