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White raven [17]
3 years ago
9

A boy throws a ball vertically up and catches it after 3 s. What height did the ball reach?

Physics
2 answers:
Crazy boy [7]3 years ago
7 0

Answer:

11.025 meters

Explanation:

The gravity’s acceleration on the surface of earth is approximately 9.8m/s

If it took 3 s to go up and come back, that means it took (1/2) × 3 to go from his hand to its highest point, and that amount of time again (1.5 seconds) to go back down to his hand.

This means the ball accelerated from Vi (initial Velocity) to 0 in 1.5 s

To find g in 1.5 i.e speed =

1.5 × 9.8 m/s = 14.7 m/s.

To find the height the ball reached=

We will use d = rt for this. Distance = Rate × Time

Since, it went 14.7 m/s at the start, and 0 m/s at the top, so the average speed of the ball going upward was (0+14.7)/2 = 7.35 m/s

7.35 m/s was the average speed = rate.

Time was 1.5 seconds, so we multiply.

7.35 m/s × 1.5seconds = 11.025 meters

Darya [45]3 years ago
3 0
The top height the ball had is (9/8)*g
g being the strength of the earth's attraction. so if g is about 9.8 then the ball went up about 11 meters
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m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

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