Question

1) How fast should you throw an apple straight down from 40m up so that its impact speed would be the same as a mango's dropped from 60m ?

2) A car accelerates uniformly at 1.5m/s2 from rest 22m/s. Then the brakes are applied and it stops 2.5s later. Find the total distance travelled.

Answer 1

Answer on Question #74492, Physics / Molecular Physics | Thermodynamics

Question. 1) How fast should you throw an apple straight down from $40\,m$ up so that its impact speed would be the same as a mango's dropped from $60\,m$?

2) A car accelerates uniformly at $1.5\,m/s^2$ from rest to $22\,m/s$. Then the brakes are applied and it stops $2.5\,s$ later. Find the total distance travelled.

Given. $h_1 = 40\,m; h_2 = 60\,m; a = 1.5\,m/s^2; v = 22\,m/s; t = 2.5\,s$.

Find. 1) $u$-? and 2) $s$-?

Solution.

1) For an apple

$$2h_1g = v^2 - u^2$$

For a mango

$$2h_2g = v^2$$

We have

$$2h_2g - 2h_1g = v^2 - v^2 + u^2 \rightarrow 2g(h_2 - h_1) = u^2 \rightarrow$$ $$u = \sqrt{2g(h_2 - h_1)} = \sqrt{2 \cdot 9.8 \cdot (60 - 40)} = 19.8\,m/s$$

Answer. $u = 19.8\,m/s$.

2) The total distance travelled

$$s = s_1 + s_2$$

For $s_1$

$$s_1 = \frac{v^2 - v_0^2}{2a} = \frac{v^2}{2a} = \frac{22^2}{2 \cdot 1.5} = 161.3\,m.$$

For $s_2$

$$s_2 = v_0 t - \frac{a t^2}{2}$$ $$v = v_0 - a t \rightarrow v_0 = a t \rightarrow a = \frac{v_0}{t} = \frac{22}{2.5} = 8.8\,m/s^2$$

So,

$$s_2 = v_0 t - \frac{a t^2}{2} = 22 \cdot 2.5 - \frac{8.8 \cdot 2.5^2}{2} = 27.5\,m$$ $$s = s_1 + s_2 = 161.3 + 27.5 = 188.8\,m$$

Answer. $s = 188.8\,m$.

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