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Exercise 11.4 · Q4

Q.A ball is thrown vertically upward from the ground with an initial velocity of 39.239.2 m/sec. If the only force considered is that attributed to the acceleration due to gravity, find:

(i) how long will it take for the ball to strike the ground?
(ii) the speed with which it will strike the ground?
(iii) how high the ball will rise?
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With upward taken as positive, the ball's acceleration is −g=−9.8 m/s2-g=-9.8\text{ m/s}^2 (gravity acts downward, opposing the upward throw). Integrate once to get velocity v(t)v(t) (using the initial speed), integrate again to get position s(t)s(t) (using s(0)=0s(0)=0 at the ground), then read off the three required quantities from v(t)v(t) and s(t)s(t).

Step 1. Set up the acceleration equation. Taking g=9.8 m/s2g=9.8\text{ m/s}^2 and upward as positive,

a=dvdt=−9.8.a = \frac{dv}{dt} = -9.8.

Step 2. Integrate to find v(t)v(t).

v=∫(−9.8) dt=−9.8t+c1.v = \int (-9.8)\,dt = -9.8t+c_1.

At t=0t=0 (the instant of throwing), v=39.2v=39.2, so c1=39.2c_1=39.2, giving

v(t)=39.2−9.8t.v(t) = 39.2-9.8t.

Step 3. Integrate to find s(t)s(t).

s=∫(39.2−9.8t) dt=39.2t−4.9t2+c2.s = \int (39.2-9.8t)\,dt = 39.2t-4.9t^2+c_2.

Measuring ss from the ground at t=0t=0, s(0)=0⇒c2=0s(0)=0 \Rightarrow c_2=0, so

s(t)=39.2t−4.9t2.s(t) = 39.2t-4.9t^2.

Step 4. (i) Find when the ball strikes the ground. The ball is back at the ground when s=0s=0 (besides the start):

39.2t−4.9t2=0⇒t(39.2−4.9t)=0⇒t=0 or t=39.24.9=8.39.2t-4.9t^2=0 \Rightarrow t(39.2-4.9t)=0 \Rightarrow t=0 \text{ or } t=\frac{39.2}{4.9}=8.

So the ball strikes the ground after t=8t=8 seconds.

Step 5. (ii) Find the speed on impact. Substitute t=8t=8 into v(t)v(t):

v(8)=39.2−9.8(8)=39.2−78.4=−39.2 m/s.v(8) = 39.2-9.8(8) = 39.2-78.4=-39.2 \text{ m/s}. …

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