Answer:
12 times
Step-by-step explanation:
The ball rebounds to 150% of its previous height each time it bounces.
To determine the height of the ball after each bounce, we multiply the preceding height by the constant 1.5. Therefore, to determine the height of the ball after n bounces we can model the scenario as a geometric sequence.
The general form of the nth term of a geometric sequence is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General form of the $n$th term of an geometric sequence}}\\\\a_n=ar^{n-1}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a_n$ is the $n$th term.}\\\phantom{ww}\bullet \;\textsf{$a$ is the first term.}\\\phantom{ww}\bullet\;\textsf{$r$ is the common ratio.}\\\phantom{ww}\bullet\;\textsf{$n$ is the position of the term.}\end{array}}[/tex]
In this case, the first term (a) is the height of the ball after the first bounce. Given that the ball is dropped from a height of 10 meters, the height of the first bounce is 150% of 10 metres, so:
[tex]a=10 \times 150\% \\\\a= 10 \times 1.5\\\\a=15[/tex]
We have already established that the common ratio (r) is 1.5. Therefore, the equation for the height of the ball after the nth bounce is:
[tex]a_n=15(1.5)^{n-1}[/tex]
To determine how many times the ball will bounce before the vertical distance of the ball exceeds 1000 m, set aₙ to greater than 1000 and solve for n:
[tex]15(1.5)^{n-1} > 1000\\\\\\(1.5)^{n-1} > \dfrac{1000}{15}\\\\\\\dfrac{1.5^n}{1.5^1} > \dfrac{1000}{15}\\\\\\1.5^n > 100\\\\\\\ln (1.5^n) > \ln(100)\\\\\\n\ln(1.5) > \ln(100)\\\\\\n > \dfrac{\ln(100)}{\ln(1.5)}}}}\\\\\\n > 11.357747174...[/tex]
As n is a positive integer, n = 12.
Therefore, the ball will bounce 12 times before the vertical distance of the ball exceeds 1000 meters.
[tex]\dotfill[/tex]
To check this, we can substitute n = 11 and n = 12 into the nth-term equation.
[tex]a_{11}=15(1.5)^{11-1}\\\\a_{11}=15(1.5)^{10}\\\\a_{11}=15(57.6650390625)\\\\a_{11}=864.9755859375\\\\a_{11}\approx 864.98\; \sf m[/tex]
So, after the 11th bounce, the ball reaches a height of approximately 864.98 meters, which is less than 1000 meters.
[tex]a_{12}=15(1.5)^{12-1}\\\\a_{12}=15(1.5)^{11}\\\\a_{12}=15(86.49755859375)\\\\a_{12}=1297.46337890625\\\\a_{12}\approx 1297.46\; \sf m[/tex]
After the 12th bounce, the ball reaches a height of approximately 1,297.46 meters, which is greater than 1000 meters.
Therefore, this proves that the ball will bounce 12 times before its vertical height exceeds 1000 meters.