Exponential distribution with rocket simulation I've come across the following question and I'm not sure how to go about solving it:

Rockets are launched, and the particles descend with acceleration due to gravity. Height/altitude can be found using $h(t) = 200 - (1/2)9.8t^2 + 8t$. Particle burn time has an exponential distribution: $B\sim \operatorname{Exp}(\lambda)$. $B$ is measured in seconds, and on average a particle burns for $2$ seconds. 

Simulate the launching of many rockets with the objective of determining how often a particle reaches ground before burning out. Assume the result for one particle for each rocket simulates the result for many.

So far I have: $2 = \operatorname{E}(B) = 1/\lambda$, so $\lambda = 1/2$ and particles reach the ground after $7.25703$ seconds. I get that the $P(B = 7.25703) = 0.01327795\ldots$

Are the above calculations correct? I'm unsure how to continue. I'm trying to simulate this using R (using only stuff like dexp(), pexp(), qexp(), rexp()).
I think the probability required is actually $P(B&gt;7.25703)=\operatorname{exp}(-\frac{1}{2}(7.25703))=0.0266338\approx0.0266$.

Exponential distribution is continuous so $P(B=t)=0$ for all $t&gt;0$.

(Just curious, if you want a higher precision for your answer maybe your gravitational constant should be of higher precision too like $g\approx9.807$.)
I think the probability required is actually $P(B&gt;7.25703)=\operatorname{exp}(-\frac{1}{2}(7.25703))=0.0266338\approx0.0266$.

Exponential distribution is continuous so $P(B=t)=0$ for all $t&gt;0$.

(Just curious, if you want a higher precision for your answer maybe your gravitational constant should be of higher precision too like $g\approx9.807$.)